As pedrosuavo suggests, the key here is conditional expectation; and, more specifically, the formula
$$\tag{1}
\DeclareMathOperator{\E}{\mathbb{E}}\E\bigl[\E[e^{tS_N}\mid N]\bigr]=\E[e^{tS_N}].
$$
Conitioned on $N=n$, $S_N$ is distributed as $S_n$, so that
$$
\E[e^{tS_N}\mid N=n]=\E[e^{tS_n}\mid N=n]=\E[e^{tS_n}],
$$
since $S_n$ and $N$ are independent. But, since $X_1,\ldots,X_n$ are independent and identically distributed, we have
$$
\E[e^{tS_n}]=\E[e^{tX_1}\cdots e^{tX_n}]=\prod_{i=1}^{n}\E[e^{tX_i}]=(\E[e^{tX_1}])^n
$$
provided $n\geq 1$. This being the case, we know that
$$
\E[e^{tX_1}]=\int_0^{\infty}\lambda e^{-\lambda x}e^{tx}\,dx.
$$
Thus the function $t\mapsto \E[e^{tX_1}]$ has domain $t\in(-\infty,\lambda)$, and for such $t$ we have $\E[e^{tX_1}]=\frac{\lambda}{\lambda-t}$.
So, for $n\geq 1$, and $t<\lambda$, we have
$$
\E[e^{tS_N}\mid N=n]=\left(\frac{\lambda}{\lambda-t}\right)^n.
$$
What about when $n=0$? We have $S_0=0$, so that $\E[e^{tS_0}]=1=\left(\frac{\lambda}{\lambda-t}\right)^0$.
So, going back to the formula (1), we have that for $t<\lambda$,
$$
\begin{align}
\E[e^{tS_N}]&=\sum_{n=0}^{\infty}\E[e^{tS_N}\mid N=n]\cdot P(N=n)\\
&=\sum_{n=0}^{\infty}\left(\frac{\lambda}{\lambda-t}\right)^n\frac{e^{-\lambda}\lambda^n}{n!}\\
&=\exp\left[\frac{t\lambda}{\lambda-t}\right].
\end{align}
$$
Based on this, we can compute $\E[S_N]$; if $\phi(x)$ is the MGF for $S_N$, then we know that $\E[S_N]=\phi'(0)=1$.
(Conveniently, this matches our heuristic idea that $\E[S_N]$ should be something like $\E[N]\cdot\E[X_i]=\lambda\cdot\lambda^{-1}$.)
