Okay, I've seen that the proof of an affirmative answer to both questions in the general case (with $\nu=\mu \otimes \lambda$) is not very hard:
Since $X \times Y$ is compact, we can let $(k_n)$ be a strictly increasing sequence of positive integers such that $\frac{1}{k_n} \sum_{i=0}^{k_n-1} f^i_\ast(\mu \otimes \lambda)$ is weakly convergent as $n \to \infty$. Let $\mathbb{P}$ denote the limit. As in the usual proof of the Krylov-Bogolyubov theorem, $\mathbb{P}$ is $f$-invariant. It remains to show that $\pi_{X\ast}\mathbb{P}=\mu$. For any continuous function $g \colon X \to \mathbb{R}$, we have \begin{align*} \int_X g \, d(\pi_{X\ast}\mathbb{P}) &= \int_{X \times Y} g \circ \pi_X \, d\mathbb{P} \\ &= \lim_{n \to \infty} \frac{1}{k_n} \sum_{i=0}^{k_n-1} \int_{X \times Y} g \circ \pi_X \, d(f^i_\ast(\mu \otimes \lambda)) \\ &= \lim_{n \to \infty} \frac{1}{k_n} \sum_{i=0}^{k_n-1} \int_{X \times Y} g(\pi_X(f^i(x,y))) \, (\mu \otimes \lambda)(d(x,y)) \\ &= \lim_{n \to \infty} \int_{X \times Y} \frac{1}{k_n} \sum_{i=0}^{k_n-1} g(\pi_X(f^i(x,y))) \, (\mu \otimes \lambda)(d(x,y)) \\ &= \int_{X \times Y} \int_X g \, d\mu \, (\mu \otimes \lambda)(d(x,y)) \\ &\hspace{20mm} \text{by Dom. Conv. Thm. and assumption on $\mu$ and $\lambda$} \\ &= \int_X g \, d\mu. \end{align*}