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Write $u_k$ for the quantity denoted ${\xi }_k$ in the question, to distinguish it from the random-walk increments. Absorption at zero at time zero gives $u_0=1$, and starting at $N$ gives $u_N=0$. For $1\le k\le N-1$, condition on the first step and use independence of the remaining increments. That step contributes one factor of $s$, so the discounted gambler's ruin hitting probability satisfiesThe absorption time is finite almost surely: in each block of $N$ steps there is a probability bounded below by a positive constant of reaching a boundary by successive steps in one direction.

Trying $u_k=r^k$ gives $ps{r}^2-r+qs=0$, with rootsThey are positive and distinct because $4pq\le 1$ and $0<s<1$. Hence the general solution is $A{r}_{+}^k+B{r}_{-}^k$. Enforcing both boundary values givesThis includes the unbiased walk without a special limiting case, since the discount keeps the roots distinct. For uniqueness, a difference of two solutions has zero boundary values and its largest absolute value $M$ satisfies $M\le sM$ if attained in the interior. Since $s<1$, $M=0$.