Showing $S$-invariance is not too hard. Note $$\int_{X \times G} f(z)d(\mu \times \nu) (z) = \int_X \int_G f(S(x,g)) d\nu(g) d\mu(x) = \int_X \int_G f(Tx,gh(x)) d\nu(g) d\mu(x)$$ $$ = \int_X \int_G f(Tx,g)d\nu(g) d\mu(x) = \int_X \int_G f(x,g)d \nu(g) d\mu(x) = \int_{X \times G} f(z)d(\mu \times \nu)(z)$$ where the third and fourth equalities used that $\nu$ is a Haar measure and $\mu$ is $T$-invariant, respectively. In general, it's easier to work with the integral characterization of being measure-preserving rather than the preimage definition.
I'm not sure ergodicity is true. Consider the case where $h(x) = e$ for all $x \in X$, where $e$ is the identity element of $G$. Note $S^n(x,g) = (T^n x, g h(T^{n-1}x)) = (T^n x, g)$. Then, for any $f_1,f_2 \in L^\infty(X\times G)$, $$\frac{1}{N}\sum_{n \le N} \int_{X \times G} f_1(z) (S^n f_2)(z)d(\mu \times \nu)(z) = \frac{1}{N}\sum_{n \le N} \int_G \int_X f_1(x,g)f_2(T^nx,g) d\mu(x)d\nu(g).$$ Now using the fact that $\frac{1}{N} \sum_{n \le N} \int_X f_1(x,g) f_2(T^n x,g) d\mu(x) \to (\int_X f_1(x,g)d\mu(x))(\int_X f_2(x,g) d\mu(x))$ for each $g$ (by ergodicity of $T$), by the Dominated Convergence Theorem, we see that $$\frac{1}{N}\sum_{n \le N} \int_{X \times G} f_1(z) (S^n f_2)(z)d(\mu \times \nu)(z) \to \int_G \left[\int_X f_1(x,g)d\mu(x)\right]\left[\int_X f_2(x,g)d\mu(x)\right] d\nu(g).$$ If $S$ were ergodic, then we would need $$\frac{1}{N}\sum_{n \le N} \int_{X \times G} f_1(z) (S^n f_2)(z)d(\mu \times \nu)(z) \to \left [\int_G \int_X f_1(x,g)d\mu(x)d\nu(g)\right] \left[\int_G \int_X f_2(x,g)d\mu(x)d\nu(g)\right].$$ But I see no reason why these last two RHS expressions should be equal. It's probably easy to come up with an example explicitly to show it need not be true, but I don't feel like doing that now.
ADDED LATER: Of course $\mu \times \nu$ is not necessarily ergodic if $h(x) = e$ for all $x \in X$. Just note that, for each $K \subseteq G$, $S^{-1}(X\times K) = X \times K$. So as long as there is some $K$ for which $\nu(K) \not = 0,1$, we're good.
