Impulse response from Frequency response: why using $e^{j\omega}$ as an input?
I think you misunderstand the notation. Writing $H(e^{j\omega})$ does not mean that $e^{j\omega}$ is the input signal. It just means that the frequency response is a function of the complex variable $e^{j\omega}$, because, as you might know, it is defined by

$$H(e^{j\omega})=\sum_{n=-\infty}^{\infty}h[n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}h[n]\big(e^{j\omega}\big)^{-n}\tag{1}$$

where $h[n]$ is the impulse response. Eq. $(1)$ is the discrete-time Fourier transform (DTFT) of the sequence $h[n]$. Generalizing to a complex argument $z$ you get the definition of the $\mathcal{Z}$-transform:

$$H(z)=\sum_{n=-\infty}^{\infty}h[n]z^{-n}\tag{2}$$

Comparing $(1)$ and $(2)$ shows that the DTFT is equal to the $\mathcal{Z}$-transform evaluated on the unit circle, i.e., for $z=e^{j\omega}$. Note that evaluating $(2)$ on the unit circle only makes sense if the unit circle is inside the region of convergence of $(2)$, otherwise $H(z)$ doesn't converge for $|z|=1$ and in that case the DTFT either doesn't exist or it takes a form which is different from $H(z)$ with $|z|=1$.

Note that some people just use $\omega$ as the argument of the function that expresses the DTFT, so you might as well come across the notation $H(\omega)$. After all, it's just a matter of convention.

Concerning the notation $H(e^{j\omega})$ you can also refer to this question and its answer.