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*}

QED.
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**An interesting fact:** I realise that there does not necessarily exist an *$f$-ergodic* invariant probability measure on $X \times Y$ whose $X$-marginal is $\mu$.

For example, take: $X=\mathbb{R}/\mathbb{Z}$; $Y=\{n^{-\frac{1}{2}} : n \in \mathbb{Z}_{\geq 1}\} \cup \{0\}$; $\mu=\mathrm{Lebesgue}$; $\lambda$ with $\lambda(\{y\})=0 \Leftrightarrow y=0$; and
$$ f([x],y) = ([x+y] , \tilde{f}(y)) $$
for $x \in \mathbb{R}$ and $y \in Y$, where
\begin{align*}
\tilde{f}(n^{-\frac{1}{2}}) &= (n+1)^{-\frac{1}{2}} \quad \forall n \in \mathbb{Z}_{\geq 1} \\
\tilde{f}(0) &= 0.
\end{align*}