As stated, this isn't true in general, even if $\{\mu_n\}$ is a constant sequence (weak convergence is a red herring).

Let $S = \mathbb{R}$ and let $\mu_n$ and $\mu$ all equal Lebesgue measure $m$.  Choose your favorite $\phi \in C_c(\mathbb{R})$ with, say, $\int_{\mathbb{R}} \phi(y)\,dy = 1$.  Set
$$h(x,y) = \begin{cases} \phi\left(y-\frac{1}{x}\right), & x > 0 \\ 0, & x \le 0. \end{cases}$$
It's easy to verify that $h$ is bounded and continuous (keep in mind that $\phi$ has compact support).  Yet we have $\int_{\mathbb{R}} h(x,y)\,dy = 1$ for all $x > 0$ and $\int_{\mathbb{R}} h(0,y)\,dy = 0$.

It will be true if $S$ is compact, since in that case you have $h(x_n, \cdot) \to h(x, \cdot)$ uniformly, and all you need to get the desired statement is the triangle inequality.