Let $(E,\mathcal E,\lambda)$ be a probability space and $\lambda$ be a measurable map on $(E,\mathcal E)$ with $\lambda\circ\tau^{-1}=\lambda$.
I would like to show that $\tau$ is $\lambda$-ergodic by reducing this claim to the following well-known result: If $(E,\mathcal E)$ is a measurable space and $(Y_i)_{i\in\mathbb N}$ is an independent stationary $(E,\mathcal E)$-valued process on any probability space (we denote the law of $Y_i$ by $\mathcal L(Y_i)$ for $i\in\mathbb N$), then the coordinate process $(X_i)_{i\in\mathbb N_0}$ on $(\Omega,\mathcal A,\operatorname P):=(E^{\mathbb N},\mathcal B(E)^{\otimes\mathbb N},\bigotimes_{i\in\mathbb N}\mathcal L(Y_i))$ is independent and stationary. Thus, the shift $$\theta:\Omega\to\Omega\;,\;\;\;(x_i)_{i\in\mathbb N}\mapsto(x_{i+1})_{i\in\mathbb N}$$ is $\operatorname P$-ergodic.
Now assume $$\tau^i=\varphi\circ\theta^iY\;\;\;\text{for all }i\in\mathbb N.\tag1$$ for some $(\mathcal A,\mathcal E)$-measurable $\varphi:\Omega\to E$ and an $(E,\mathcal E)$-valued independent stationary process $(Y_i)_{i\in\mathbb N}$ on $(E,\mathcal E,\lambda)$.
Can we somehow use the identity $(1)$ to derive that $\tau$ is $\lambda$-ergodic from knowing that $\theta$ is $\operatorname P$-ergodic? I guess we only need to show that $\varphi$ is sufficiently nice for this conclusion to hold.
Feel free to assume that $(Y_i)_{i\in\mathbb N}$ is identically distributed, if necessary.
Remark: I've asked for the special case, where $(Y_i)_{i\in\mathbb N}$ is the dynamical system generated by the Bernoulli shift, back over on MSE: https://math.stackexchange.com/q/3657914/47771.