Output of lowpass filter with damped sine wave input
I've finally figured this out thanks to a friend of mine. The thing is that $N(t)$ can be even or odd, and in both cases, the signal we get is easy to deal with.

We can calculate the desired expectation this way:

$$\mathbb{E}[Z(t)]=\mathbb{E}[h(t)*Y(t)]=\mathbb{E}[h(t)*Y(t)|N(t) =2n]\mathbb{P}[N(t)=2n]+\mathbb{E}[h(t)*Y(t)|N(t) =2n+1]\mathbb{P}[N(t)=2n+1]$$

But when $N(t)$ is even or odd, $Y(t)$ becomes simple:

$$Y(t) =
\left\{
 \begin{array}{ll}
  \cos^2(\omega_0t)  &amp; \mbox{if } N(t)\mbox{ is even } \\
  -\cos(\omega_0t)\sin(\omega_0t)    &amp; \mbox{if } N(t)\mbox{ is odd }
 \end{array}
\right.$$

So

$$\mathbb{E}[Z(t)]=\mathbb{P}[N(t)=2n]\cdot h(t)*\cos^2(\omega_0t)-\mathbb{P}[N(t)=2n+1]\cdot h(t)*(\cos(\omega_0t)\sin(\omega_0t))$$

And those two probabilities are easy to calculate, as well as the two convolutions left.