Finding the output of a system where the input is a sum of complex exponentials
Okay... I managed to do it on my own. Here's the solution if you're interested. 

We know that $$H\{x(t)\} = \sum_{k=0}^{\infty}a_k \  H\{e^{jw_kt}\}$$ since $H$ is a linear system. 

We also know that $H\{e^{jw_kt}\}  = F(w_k) \ e^{jw_kt}$, where $F(w_k)$ is the frequency response of the system. 

By definition, we have that $$F(w_k) = \int_{-\infty}^{+\infty}h(t) \ e^{-jw_kt} \ dt \\ = \frac{1}{\tau} \left(\int_{-\infty}^{+\infty}\delta(t+\tau)e^{-jw_kt} dt \ - \ \int_{-\infty}^{+\infty}\delta(t) e^{-jw_kt} dt\right)\\ =\frac{1}{\tau} \left( \int_{-\infty}^{+\infty} \delta(u) e^{\ jw_k(\tau-u)} \ du \ - \ 
\int_{-\infty}^{+\infty} \delta(t)e^{\ jw_k(0-t)}\ dt \right) \\ = \frac{e^{ \ jw_k\tau} - e^0}{\tau}$$

Therefore we can simply write $$H\{x(t)\} = \sum_{k=0}^{\infty} \frac{e^{\ jw_k\tau}-1}{\tau} \ a_ke^{\ jw_kt}$$