(EDITED, to comply with quite accurate objections by Louigi and Byron)
Assume that with full probability $X_k$ is either $-1$ or a random positive integer (this includes the setting of the question when $p=1/(k+1)$ with $k$ a positive integer but note that $X_k$ may take more than one positive integer values). Then, 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 $$ E(S_N)=E(X+X^2;X>0)/(2q). $$ If $X=-1$ or $X=k$ with $k$ a positive integer, then $[X > 0]=[X=k]$ and $p=1/(k+1)$, and one sees that $S_N$ is uniformly distributed on the integers from $1$ to $k$ and that $E(S_N)=(k+1)/2$.