$$ x_\mathrm{new}(t) = I_\mathrm{new}(t)\cdot \cos(2\pi f_0 t) - Q_\mathrm{new}(t)\cdot \sin(2\pi f_0 t) $$

By use of this approach you can receive the signal at the carrier frequency $f_c+f_\Delta = 500\:$MHz, perform quadrature down-conversion as normal, and the apply the above technique to modify the effective carrier frequency by using e.g. $f_\Delta = 100\:$Hz or whatever. As shown above you do not need to demodulate the signal and the conversion can be applied to any signal and be handled at baseband.

One thing to remember, though, is that the requirement to sampling frequency increases as the bandwidth of the "new" signal is wider than the original signal. In case of a $100\:$Hz offset this is likely not a big issue but it obviously depends on the bandwidth $B$ of the original baseband signal. The conversions above do not require expensive and possibly distorting filters.

$$ x(t) = I_\mathrm{new}(t)\cdot \cos(2\pi f_0 t) - Q_\mathrm{new}(t)\cdot \sin(2\pi f_0 t) $$

By use of this approach you can receive the signal at the carrier frequency $f_c+f_\Delta = 500\:$MHz, perform quadrature down-conversion as normal, and the apply the above technique to modify the effective carrier frequency by using e.g. $f_\Delta = 100\:$Hz or whatever. As shown above you do not need to demodulate the signal and the conversion can be applied to any signal and be handled at baseband. Depending on the actual application, which is not known from the question, it may not be allowed to use the above. You can extract the original $I$/$Q$ signals directly from $x(t)$ if this signal is quadrature down-converted with the carrier frequency $f_c$. If you quadrature down-convert $x(t)$ with a locally generated frequency of $f_0$ you extract $I_\mathrm{new}(t)$/$Q_\mathrm{new}(t)$. If you have $I_\mathrm{new}(t)$/$Q_\mathrm{new}(t)$ you can reconstruct $I(t)$/$Q/t)$ from the above equations (solving two equations with two unknowns).

One thing to remember, though, is that the requirement to sampling frequency increases as the bandwidth of the "new" signal is wider than the original signal. In case of a $100\:$Hz offset this is likely not a big issue but it obviously depends on the bandwidth $B$ of the original baseband signal.

If the problem is more in the direction of RX a signal at one carrier and relaying to TX the same signal at another frequency then some other technique should be applied. An obvious technique is to quadrature down-convert and then just quadrature up-convert to the desired offset carrier. Also a translational loop combined with amplitude information can be applied. The translational loop filters the phase part of the signal. If further filtering is needed depends entirely on the required spec.

Perhaps I reply too late but never the less ..

I am not sure how this problem is actually intended to be understood and the info on electronics.SE does not help much. Of questions some could be:

1. Why demodulate at the carrier? Should it be understood in the way that you want to convert the information at the carrier to baseband in-phase ($I$) and quadrature-phase ($Q$) signals?
2. What do you mean by the original signal? Is that an RF signal or the $I$/$Q$ signal components?
3. Why do you want to modulate (I assume you just mean frequency translate) the signal to the new carrier frequency?
4. What is actually the objective of the question?

Without this information I am forced to guess a bit.

As I understand it you have a QAM (for the following, the modulation is not relevant) signal at a carrier frequency $f_c=$500$\:$MHz which can be expressed as:

$$ x(t) = I(t)\cdot \cos(2\pi f_c t) - Q(t)\cdot \sin(2\pi f_c t) $$

where the in-phase signal $I(t)$ occupies the frequency band $-B \leq f \leq B$ and the quadrature-phase signal $Q(t)$ also occupies the frequency band $-B \leq f \leq B$.

To simplify things I have omitted an initial constant phase in $\cos()$/$\sin()$ - this can easily be included though and in fact any constant phase can be included in $I$ and $Q$ without loss of generality.

To me it seems as if you want to force an offset by saying that the carrier is $f_c=f_0+f_\Delta$ and then you would like to know how "new" $I$/$Q$ signals referred to just the carrier frequency $f_0$ looks like. This means that the "new" RF signal is:

$$ x_\mathrm{new}(t) = I(t)\cdot \cos(2\pi (f_0+f_\Delta) t) - Q(t)\cdot \sin(2\pi (f_0+f_\Delta) t) $$

which we need to rewrite as:

$$ x_\mathrm{new}(t) = I_\mathrm{new}(t)\cdot \cos(2\pi f_0 t) - Q_\mathrm{new}(t)\cdot \sin(2\pi f_0 t) $$

This modification be done by applying standard trigonometric identities (see here). What we need is:

$$ \cos(\alpha + \beta) = \cos(\alpha)\cdot\cos(\beta) - \sin(\alpha)\cdot\sin(\beta) $$

$$ \sin(\alpha + \beta) = \sin(\alpha)\cdot\cos(\beta) + \cos(\alpha)\cdot\sin(\beta) $$

Using these identities we can then determine the "new" in-phase and quadrature-phase signal components as:

$$ I_\mathrm{new}(t) = I(t) \cdot \cos(2\pi f_\Delta t) - Q(t) \cdot \sin(2\pi f_\Delta t) $$

$$ Q_\mathrm{new}(t) = I(t) \cdot \sin(2\pi f_\Delta t) + Q(t) \cdot \cos(2\pi f_\Delta t) $$

The "new" in-phase and quadrature-phase baseband signals occupy the frequency band $-B + \vert f_\Delta \vert \leq f \leq B + \vert f_\Delta \vert$. The bandwidth increases no matter if we change $f_\Delta$ in the positive or negative direction. Should it be needed the opposite transformation can also be done such that move from known "new" signals $I_\mathrm{new}$/$Q_\mathrm{new}$ to the $I$/$Q$ signals.

By use of this approach you can receive the signal at the carrier frequency $f_c+f_\Delta = 500\:$MHz, perform quadrature down-conversion as normal, and the apply the above technique to modify the effective carrier frequency by using e.g. $f_\Delta = 100\:$Hz or whatever. As shown above you do not need to demodulate the signal and the conversion can be applied to any signal and be handled at baseband.

One thing to remember, though, is that the requirement to sampling frequency increases as the bandwidth of the "new" signal is wider than the original signal. In case of a $100\:$Hz offset this is likely not a big issue but it obviously depends on the bandwidth $B$ of the original baseband signal. The conversions above do not require expensive and possibly distorting filters.

Perhaps I reply too late but never the less ..

I am not sure how this problem is actually intended to be understood and the info on electronics.SE does not help much. Of questions some could be:

1. Why demodulate at the carrier? Should it be understood in the way that you want to convert the information at the carrier to baseband in-phase ($I$) and quadrature-phase ($Q$) signals?
2. What do you mean by the original signal? Is that an RF signal or the $I$/$Q$ signal components?
3. Why do you want to modulate (I assume you just mean frequency translate) the signal to the new carrier frequency?
4. What is actually the objective of the question?

Without this information I am forced to guess a bit.

As I understand it you have a QAM (for the following, the modulation is not relevant) signal at a carrier frequency $f_c=$500$\:$MHz which can be expressed as:

$$ x(t) = I(t)\cdot \cos(2\pi f_c t) - Q(t)\cdot \sin(2\pi f_c t) $$

where the in-phase signal $I(t)$ occupies the frequency band $-B \leq f \leq B$ and the quadrature-phase signal $Q(t)$ also occupies the frequency band $-B \leq f \leq B$.

To simplify things I have omitted an initial constant phase in $\cos()$/$\sin()$ - this can easily be included though and in fact any constant phase can be included in $I$ and $Q$ without loss of generality.

To me it seems as if you want to force an offset by saying that the carrier is $f_c=f_0+f_\Delta$ and then you would like to know how "new" $I$/$Q$ signals referred to just the carrier frequency $f_0$ looks like. This means that we can rewrite the original signal as:

$$ x(t) = I(t)\cdot \cos(2\pi (f_0+f_\Delta) t) - Q(t)\cdot \sin(2\pi (f_0+f_\Delta) t) $$

which we need to rewrite as:

$$ x_\mathrm{new}(t) = I_\mathrm{new}(t)\cdot \cos(2\pi f_0 t) - Q_\mathrm{new}(t)\cdot \sin(2\pi f_0 t) $$

This modification be done by applying standard trigonometric identities (see here). What we need is:

$$ \cos(\alpha + \beta) = \cos(\alpha)\cdot\cos(\beta) - \sin(\alpha)\cdot\sin(\beta) $$

$$ \sin(\alpha + \beta) = \sin(\alpha)\cdot\cos(\beta) + \cos(\alpha)\cdot\sin(\beta) $$

Using these identities we can then determine the "new" in-phase and quadrature-phase signal components as:

$$ I_\mathrm{new}(t) = I(t) \cdot \cos(2\pi f_\Delta t) - Q(t) \cdot \sin(2\pi f_\Delta t) $$

$$ Q_\mathrm{new}(t) = I(t) \cdot \sin(2\pi f_\Delta t) + Q(t) \cdot \cos(2\pi f_\Delta t) $$

The "new" in-phase and quadrature-phase baseband signals occupy the frequency band $-B + \vert f_\Delta \vert \leq f \leq B + \vert f_\Delta \vert$. The bandwidth increases no matter if we change $f_\Delta$ in the positive or negative direction. Should it be needed the opposite transformation can also be done such that move from known "new" signals $I_\mathrm{new}$/$Q_\mathrm{new}$ to the $I$/$Q$ signals.

By use of this approach you can receive the signal at the carrier frequency $f_c+f_\Delta = 500\:$MHz, perform quadrature down-conversion as normal, and the apply the above technique to modify the effective carrier frequency by using e.g. $f_\Delta = 100\:$Hz or whatever. As shown above you do not need to demodulate the signal and the conversion can be applied to any signal and be handled at baseband.

One thing to remember, though, is that the requirement to sampling frequency increases as the bandwidth of the "new" signal is wider than the original signal. In case of a $100\:$Hz offset this is likely not a big issue but it obviously depends on the bandwidth $B$ of the original baseband signal. The conversions above do not require expensive and possibly distorting filters.