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.

Edit: Previous answer was bogus.

Let me change your notation a little to let $x_0$ be the limit of the $x_n$.

Yes, this is true. It follows from tightness and the general fact that $h(x_n, \cdot) \to h(x_0, \cdot)$ uniformly on compact sets.

Without loss of generality, let's suppose that $h(x_0, \cdot) = 0$ (replace $h(x,y)$ by $h(x,y) - h(x_0,y)$).

Let $\epsilon > 0$ and let $M$ be the sup norm of $h$. By the easier direction of Prohorov's theorem, the sequence $\{\mu_n\}$ is tight, so there is a compact $K \subset S$ such that for every $n$ we have $\mu_n(K^C) < \epsilon$. Now we have $$\begin{align*}\left|\int h(x_n, y) \mu_n(dy)\right| &\le \int_K |h(x_n,y)|\,\mu_n(dy) + \int_{K^C} |h(x_n, y)| \mu_n(dy) \\ &\le \sup_{y \in K} |h(x_n, y)| + M \epsilon.\end{align*}$$ So it suffices to show that $h(x_n, \cdot) \to 0$ uniformly on $K$. Suppose not; then passing to a subsequence if necessary, we can find $\delta > 0$ so that $\sup_{y \in K} |h(x_n, y)| > \delta$ for all $n$. Thus for each $n$ we can find $y_n \in K$ such that $|h(x_n, y_n)| > \delta$. By compactness, we can pass to a further subsequence so that $y_n$ converges to some $y_0$. Now by continuity of $h$ we have $|h(x_n, y_n)| \to |h(x_0, y_0)| = 0$, a contradiction.

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.

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.