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A transfer function is defined as the Laplace transform of the ratio of output to input.
Also, every LTI system has an eigenfunction. Given such eigenfunction as an input, the ratio of the output to input is a constant, which means the transfer function is constant.
You're basically right: the frequency response evaluated at a single frequency point is a complex constant. Note that a better notation for the input signal would have been $x(t)=e^{j\omega_0t}$, emphasizing that $\omega_0$ is just an arbitrary but fixed frequency, and not equal to $\omega$, which is commonly used as the independent variable of the Fourier transform. So the constant you came up with is simply $H(j\omega_0)$.
Apart from that, the main problem in your "proof" is that you can't take the Laplace transform of $e^{j\omega_0t}$ because it doesn't exist. Even taking the Fourier transform - which does exist as a generalized function - doesn't help, because you get a meaningless fraction of two Dirac impulses.
Of course it is true that if the Fourier transform of the input is $\delta(\omega-\omega_0)$ (corresponding to a time-domain input $e^{j\omega_0t}$), then the Fourier transform of the output will be $H(j\omega)\delta(\omega-\omega_0)=H(j\omega_0)\delta(\omega-\omega_0)$. However, as mentioned above, you can't just divide these two frequency domain expressions in order to find the system's frequency response.
If you use a complex exponential as an input, you can divide in the time-domain in order to find the frequency response at one specific frequency: if $x(t)=e^{j\omega_0t}$, then $y(t)=H(j\omega_0)e^{j\omega_0t}$ and $H(j\omega_0)=y(t)/x(t)$, which is of course a generally complex-valued constant.
Note that in general an input signal $x(t)$ with Fourier transform $X(j\omega)$ will only allow you to determine a system's frequency response at frequencies for which $X(j\omega)\neq 0$ holds, namely by computing $Y(j\omega)/X(j\omega)$. For a complex exponential, $X(j\omega)\neq 0$ is only satisfied at a single frequency, and, consequently, you can determine the frequency response only at that single frequency.
When the input to an LTI system is an eigenfunction, then:
$$ x(t) = e^{j\omega_0 t} => y(t) = H(\omega_0) e^{j\omega_0 t}$$
where $H(\omega_0)$ is the associated eigenvalue of the LTI system; the transfer function evaluated at the particular frequency:
$$ H(\omega_0) = H(s)|_{s = j\omega_0}. $$
Here not the transfer function $H(s)$, but just one value of it is found by just one eigenfunction.
But if you use the general expression for the eigenfunction, then the input is $x(t) = e^{j\omega t}$ and the output will be $$y(t) = H(j\omega) e^{j \omega t}. $$ Here you assume that $H(j\omega) = a$ ; a constant, and therefore in the fraction expression you will find out that $H(s) = \frac{Y(s)}{X(s)} = a = H(j\omega)$. Here, the constant $a$ is actually the unknown variable $H(j\omega)$...
Note that the eigenvalue $H(\omega)$ will be a complex-constant for a given particular frequency $\omega_0$. But as the given frequency changes, the constant (in general) will change too. It's just like an ordinary function $f(x)$ treated as a single value for a particular $x = x_0$ but a rather a varying mapping for general $x$ values...
The eigenfunctions $e^{j\omega t}$, depending on parameter $\omega$, have each a specific eigenvalue $a_\omega$ (potentially different). So for any linear operator $\mathcal{L}$ (Laplace included), even formally (not considering existence), you cannot factorize the $a_\omega$ in $\mathcal{L}(\sum a_\omega e^{j\omega t})$ outside the sum, unless all $a_\omega$ are equal.
If they were equal, then you would have a constant transfer function, indeed.
To make the visualization of the "too quick jump" more concrete, i'll cast your initial reasoning for polynomials. It's a bit like: any monomial divided by the proper power of $x$ is a constant. A polynomial is a sum of monomials, hence by dividing the polynomial by the proper powers, every polynomial is a constant [which is generally false]
