Assuming that with full probability $X_k$ is either $-1$ or $>0$ (this includes the setting of the question), Wiener-Hopf factorization formula becomes simple enough to compute the distribution of $S_N$. 

More precisely, let $N$ denote the first time $n\ge1$ such that $S_n>0$ (as in the OP's post) and let $M$ denote the first time $n\ge1$ such that $S_n\le 0$ (note the "lower than or equal to"). In the centered and bounded case the OP is interested in, $N$ and $M$ are both almost surely finite and Wiener-Hopf formula reads
$$
(1-E(e^{iuS_N}))(1-E(e^{iuS_M}))=1-E(e^{iuX}),
$$
for every real number $u$ and every $X$ distributed as the steps $X_k$. Here, $S_M=-1$ on $[X_1=-1]$ and $S_M=0$ on $[X_1>0]$. This yields
$$
q(1-e^{-iu})E(e^{iuS_N})=E(e^{iuX};X>0)-p,
$$
with $q=P[X=-1]$ and $p=1-q=P[X>0]$. This provides the full distribution of $S_N$ and, differentiating both sides at $u=0$, the expectation of $S_N$. The end result is
$$
E(S_N)=E(X+X^2;X>0)/(2q).
$$
In the case the OP is interested in, $[X > 0]=[X=q/p]$, hence $E(S_N)=1/(2p)$.