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.