Beginner LTI system Fourier transform question
You're confusing time and frequency domain. Why should the (time domain) output signal $y(t)$ be a function of $\omega$? If $x(t)=e^{j\omega_0t}$ is the input signal to an LTI system with frequency response $H(j\omega)$, then the output signal is given by

$$y(t)=e^{j\omega_0t}H(j\omega_0)\tag{1}$$

Note that in $(1)$, $H(j\omega_0)$ is a constant, not a function of $\omega$.

Equation $(1)$ can be easily derived from the convolution integral

$$y(t)=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\tag{2}$$

with $x(t)=e^{j\omega_0t}$ and the Fourier transform relation

$$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt\tag{3}$$

I'm sure you can fill in the details yourself.



Your last equation would be true for the Fourier transform of the output signal:

$$Y(j\omega)=X(j\omega)H(j\omega)\tag{4}$$

And with $X(j\omega)=2\pi\delta(\omega-\omega_0)$ you obtain

$$Y(j\omega)=2\pi\delta(\omega-\omega_0)H(j\omega)=2\pi\delta(\omega-\omega_0)H(j\omega_0)\tag{5}$$

which is of course just the Fourier transform of Eq. $(1)$.