I thinks this is an application of Ito's Lemma with noncontinuous semimartingales. Of course, in the case that $X$ is a Poisson process, it simpliefies to $f(X_t)-f(X_0)=\sum_{0< s\le t} \Delta [f(X_{s})],$ where $f(x)=e^{\alpha x}$. Since the Poisson process has a jump size 1 a.s., $\sum_{0 So the process you are looking for is $b_s=f(X_s+1)-f(X_s)=\exp(\alpha X_s)(e^\alpha-1).$
With that, the last question is easy to answer, since $\int_0^t \exp(\alpha X_{s^-})dX_s=\frac1{e^{\alpha}-1}[\exp(\alpha X_t)-1],$ and you can use the moment generating function for $X_t$ $\mathbb{E} \exp(\alpha X_t)=\exp\{\lambda t (e^{\alpha}-1)\}$ to figure out expectation and variance of $\int_0^t \exp(\alpha X_{s^-})dX_s.$