Here's a probabilistic solution to your problem. Suppose $X_N$ is a binomial random variable with parameters $N$ and $s/N$ (ie. $X_N$ counts the number of heads in $N$ independent coin flips, each with probability $s/N$ of heads). Also let $P$ be a Poisson random variable with mean $s$, so that for all non-negative integers $k$:
$$Prob[P=k] = e^{-s}\frac{s^k}{k!}.$$
A well-known result sometimes called the law of rare events implies that the distribution of $X_N$ converges to that of $P$ as $N\to +\infty$. In particular, for any bounded real-valued function $f$ defined on the non-negative integers:
$$E(f(X_N)) = \sum_{k=0}^N \binom{N}{k}\left(1-\frac{s}{N}\right)^{N-k}\left(\frac{s}{N}\right)^{k}f(k)\to E(f(P))=\sum_{k=0}^{+\infty}e^{-s}\frac{s^k}{k!}f(k).$$

Apply this to $f$ satisfying $f(x)=1/x$ if $x>0$, $f(0)=0$, and the LHS becomes your expression. The RHS becomes:
$$\sum_{k=1}^{+\infty}e^{-s}\frac{s^k}{k!\times k} = e^{-s}\int_{0}^{s}\frac{e^{u}-1}{u}du,$$
which I guess is the same as the previous answer.

**Added note:** a quantitative version of the law of rare events gives the error bound:
$$\forall f:N\to [0,1], |E(f(X_N))-E(f(P))|\leq s\left(1-e^{-s/N}\right);$$
this allows for simultaneous limits in N and $s$, and goes to $0$ iff $s^2/N\to 0$.