Note the OP has mixed continuous time and discrete time formulations. My answer below is given with continuous time $t$ but can equally be applied to discrete time index $n$.
A real signal cannot contain only negative frequencies but will have a complex conjugate symmetric spectrum which means equal magnitude and opposite phase negative and positive frequency components. That said method 1 can work because of the low pass filter: to shift the negative frequency passband component to DC, we multiply it with a positive frequency only local oscillator, which is complex as $e^{j\omega_c t}= \cos(\omega_c t)+ j\sin(\omega_c t)$. This will shift the signal upward in frequency such that the negative frequency component is centered on DC (assuming it was at the frequency given by $\pm \omega_c$) and the positive frequency component will be shifted up to $2\omega_c$ and subsequently removed by the low pass filter.
Thus to do method 1 properly with $x(t)$ as the real Input we have:
$$x(t)e^{j\omega t}= x(t)\cos(\omega_c t)+ jx(t)\sin(\omega_c t)$$
Resulting in the real and imaginary components at the input to the low pass filters on each path.
The frequency spectrums showing the processing in Method 1 are depicted below:
Method 2 if done properly allows us to eliminate the low pass filtering. To be clear, the ‘Hilbert’ function returns the analytic signal as $x(t)+ j\hat{x}(t)$ which is a complex passband signal which only has positive frequency components. ($\hat{x}(t)$ is the Hilbert Transform of $x(t)$.) The waveform would still be at frequency $\omega_c$ but exist at the positive frequency component only (as a complex signal with I and Q paths). Thus we can move this signal down to baseband by multiplying with a complex LO given as $e^{-j\omega_c t}$, which would require four real multiplications (multiply out the complex input and complex LO as $(I_1+jQ_1)(I_2+jQ_2)$ to get implementation), but to note importantly that the spectrum would be mirrored from the method 1 given which moved the negative frequency component to baseband. This is of little consequence as the signal is the same just mirrored, and it is trivial to mirror the spectrum back by changing the sign of Q in the result, or alternatively change the sign of Q in the analytic signal that was returned by ‘Hilbert’ (which will then indeed be the negative frequency component), and multiply that result by $e^{j\omega t}$.
That said, the correct approach using method 2 would be:
$$(x(t)-j\hat{x}(t))(\cos(\omega_c t)+j\sin(\omega_c t))$$
Where the above terms are multiplied out to get the real and imaginary components for the output as $I_o+jQ_o$:
$$I_o(t) = x(t)\cos(\omega_c t) + \hat{x}(t)\sin(\omega_c t)$$
$$Q_o(t) = x(t) \sin(\omega_c t) - \hat{x}(t)\cos(\omega_c t)$$
The frequency spectrums showing the processing in Method 2 are depicted below:
With method 1 we only need two real multipliers but requires the additional low pass filters (may not be that bad to do for the case of a carrier significantly higher than the signal bandwidth along with high enough sampling rates to support the higher frequency component without aliasing), while with method 2 we require four real multipliers but eliminate the need for additional filtering.
As the OP pointed out to me in the comments, the filter complexity to do the low pass filter and Hilbert filter would be similar. That said the advantages of Method 2 is that we can filter once (real $x(t)$ with a real Hilbert filter, and if properly done such that the delay matched path is just an integer sample delay the processing is minimized to just one filter). This then requires four real multipliers for the like and cross terms as detailed above. Compared to method 1, the low pass filter would be of similar complexity, but we have to filter the real and imaginary legs of the complex product; the operations to do this would significantly offset the savings of just having two multipliers.
This is diagrammed below with the implementation for Method 1 where $h(t)$ represents the impulse response for the low pass filter:
Compared to that for Method 2, where here $h(t)$ is different but of similar complexity as the impulse response for the Hilbert Transform approximation:
Method 2 appears to be more complicated, but if the filter complexity was similar, Method 2 only requires half the filtering operations compared to Method 1. Below is an example plot showing how I would construct a discrete time Hilbert from a Half band filter by heterodyning the Halfband coefficients. We see from this that if the Halfband represented the required low pass filter for Method 1, the same filter complexity would be useful as the Hilbert for Method 2 (with a trivial integer sample delay for the delayed matched path).
That said, when the carrier frequency is similar to the bandwidth (where we have the upper sideband and lower sideband both in proximity to DC), Method 2 has a clear advantage in terms of lower overall processing required for a discrete-time implementation. However when the carrier is significantly higher than the bandwidth Method 1 has an advantage as even though a simpler bandpass Hilbert could be used for Method 2 in this case, the low pass filters for Method 1 can run at a lower sampling rate (more efficient decimating low pass filters can be used).
It may also be thought that for the case of discrete time processing, method 1 would require a higher sampling frequency, but this is not actually the case: both methods can operate with a Nyquist frequency just above the upper bandwidth of $x(t)$ with sufficient margin for filtering. If this were the case with method 1, the upper sideband would circularly rotate to be half in the positive frequency (straddling Nyquist) and half in the negative frequency, and the low pass filter would equivalently reject it.
