 Solution

At each index the two independent uniform symbols match with probability $p=k/k^2=1/k$. The match indicators are independent across indices. Consequently the waiting times between successive matches are independent geometric random variables on $1,2,\dots$ with success probability $p$, and $S_m$ is their sum.

For one waiting time $G$, summing the geometric series gives $\mathbb{E}{e}^{tG}=p{e}^t/[1-(1-p)e^t]$. Independence yieldsFor $k>1$ the moment-generating function is finite for $t<\log (k/(k-1))$. When $k=1$, every index matches, $S_m=m$ deterministically and the formula reduces to $e^{mt}$ for every real $t$.