I have found an answer; it is based on Proposition 3.10 of here.
Claim: $\mathbb{P}_W \otimes \lambda$ is the only $\Theta$-invariant probability measure whose projection onto $\Omega$ is $\mathbb{P}_W$.
I start by proving that $\mathbb{P}_W \otimes \lambda$ is $\Theta$-ergodic (which I was already able to prove before - however, there may well exist a nicer proof than mine):
Proof that $\mathbb{P}_W \otimes \lambda$ is ergodic. Let $P$ be the Markov transition function describing the dynamics on $\mathbb{S}^1$; that is,
$$P(x,A) \ := \ \mathbb{P}_{N(0,1)}(\alpha \in \mathbb{R}:R_\alpha(x)\in A)$$
for all $x \in \mathbb{S}^1$ and $A \in \mathcal{B}(\mathbb{S}^1)$. Obviously $\lambda$ is a stationary distribution for $P$. We also clearly have that for every $\lambda$-positive measure set $A \in \mathcal{B}(\mathbb{S}^1)$, $P(x,A)>0$ for all $x \in \mathbb{S}^1$; therefore, $\lambda$ must be the only stationary distribution, and so in particular, must be ergodic.
Consequently, $\Theta$ is an ergodic transformation of the probability space $(\Omega \times \mathbb{S}^1,\mathcal{F}_0^\infty \otimes \mathcal{B}(\mathbb{S}^1),\left.\mathbb{P}_W\right|_{\mathcal{F}_0^\infty} \otimes \lambda)$, where $\mathcal{F}_0^\infty:=\sigma(\omega \mapsto \omega(t):t \geq 0)$. (For a justification of this, see e.g. Theorem 143(ii) of my notes, which is a slight generalisation of Theorem 3.1 of here or Theorem I.2.1 of here.)
Now suppose for a contradiction that $\Theta$ is not an ergodic transformation of the full probability space $(\Omega \times \mathbb{S}^1,\mathcal{F} \otimes \Sigma, \mathbb{P}_W \otimes \lambda)$; so we can write $\mathbb{P}_W$ as a non-trivial convex combination of two distinct $\Theta$-invariant probability measures $\mu_1$ and $\mu_2$. Since $\left.\mathbb{P}_W\right|_{\mathcal{F}_0^\infty} \otimes \lambda$ is ergodic under $\Theta$, we have that
$$ \left.\mu_1\right|_{\mathcal{F}_0^\infty \otimes \mathcal{B}(\mathbb{S}^1)} \ = \ \left.\mu_2\right|_{\mathcal{F}_0^\infty \otimes \mathcal{B}(\mathbb{S}^1)} \ = \ \left.\mathbb{P}_W\right|_{\mathcal{F}_0^\infty} \otimes \lambda. $$
Now let $\mathcal{F}_T^\infty:=\sigma(\omega \mapsto \omega(t):t \geq T)$ for each $T \in \mathbb{R}$. For any $n \in \mathbb{N}$, it is clear that $\theta^n$ is $(\mathcal{F}_0^\infty,\mathcal{F}_{-n}^\infty)$-measurable and that the map $\omega \mapsto \omega(n)$ is $(\mathcal{F}_0^\infty,\mathcal{B}(\mathbb{R}))$-measurable. So then, $\Theta^n\colon(\omega,x) \mapsto (\theta^n\omega,R_{\omega(n)}(x))\,$ is $\,(\mathcal{F}_0^\infty \otimes \mathcal{B}(\mathbb{S}^1),\mathcal{F}_{-n}^\infty \otimes \mathcal{B}(\mathbb{S}^1))$-measurable. So, for any $A \in \mathcal{F}_{-n}^\infty \otimes \mathcal{B}(\mathbb{S}^1)$, we have that
$$ \mu_1(A) = \mu_1(\Theta^{-n}(A)) = \mu_2(\Theta^{-n}(A)) = \mu_2(A). $$
So then, since $\mu_1$ and $\mu_2$ agree on $\mathcal{F}_{-n}^\infty \otimes \mathcal{B}(\mathbb{S}^1)$ for all $n \in \mathbb{N}$, it follows that $\mu_1$ and $\mu_2$ agree on the whole of $\mathcal{F} \otimes \mathcal{B}(\mathbb{S}^1)$, contradicting our assumption that $\mu_1$ and $\mu_2$ are distinct measures. QED
With this, to prove the Claim, we just apply the following theorem (adapted from Proposition 3.10 of the book by Furstenberg that I cited at the top of this answer):
Theorem. Let $(\Omega,\mathcal{F})$ be a standard measurable space, let $\theta:\Omega \to \Omega$ be a measurable map, let $X$ be a compact metrisable abelian topological group with $\lambda$ denoting the Haar measure, and let $R:\Omega \to X$ be a measurable function. Define the function $\Theta:\Omega \times X \to \Omega \times X$ by $\Theta(\omega,x)=(\theta\omega,R(\omega)x)$. For any $\theta$-invariant probability measure $\mathbb{P}$, if $\mathbb{P} \otimes \lambda$ is $\Theta$-ergodic, then the only $\Theta$-invariant probability measure $\mu$ satisfying $\mu(E \times X)=\mathbb{P}(E)$ for all $E \in \mathcal{F}$ is $\,\mu=\mathbb{P} \otimes \lambda$.
Now Furstenberg actually assumes that $\theta$ is uniquely ergodic and concludes that $\Theta$ is uniquely ergodic. Nonetheless, the proof given directly yields the theorem as I have given it above. I will now write the proof of the theorem:
Proof of Theorem. Fix a compact metrisable topology on $\Omega$ whose Borel $\sigma$-algebra coincides with $\mathcal{F}$. For any probability measure $\mu$ on $\Omega \times X$ and any point $(\omega,x) \in \Omega \times X$, we will say that $(\omega,x)$ is $\mu$-generic if
$$ \frac{1}{N} \sum_{n=0}^{N-1} f(\Theta^n(\omega,x)) \ \underset{N \to \infty}{\to} \ \int_{\Omega \times X} f(\tilde{\omega},\tilde{x}) \, \mu(d(\tilde{\omega},\tilde{x})) \hspace{4mm} \forall \, f \in C(\Omega \times X,\mathbb{R}). $$
Observation 1 [c.f. Proposition 3.7 of Furstenberg]: For any $\Theta$-ergodic probability measure $\mu$, $\mu$-almost every point in $\Omega \times X$ is $\mu$-generic. [It seems to me that the continuity of the dynamical system in Proposition 3.7 is not needed; all that matters is that the relevant class of functions on the phase space admits a countable uniformly dense subset.]
To see this: Let $\mu$ be any $\Theta$-ergodic probability measure, and let $S$ be a countable dense subset of $C(\Omega \times X,\mathbb{R})$. By Birkhoff's ergodic theorem, the set $Y \subset \Omega \times X$ of points $\mathbf{x}$ with the property that
$$ \frac{1}{N} \sum_{n=0}^{N-1} f(\Theta^n(\mathbf{x})) \ \underset{N \to \infty}{\to} \ \int_{\Omega \times X} f(\tilde{\mathbf{x}}) \, \mu(d\tilde{\mathbf{x}}) \hspace{4mm} \forall \, f \in S $$
is a $\mu$-full measure set. We now show that every point in $Y$ is $\mu$-generic. Fix any $\mathbf{x} \in Y$, any $f \in C(\Omega \times X,\mathbb{R})$ and any $\varepsilon>0$. Pick any $f' \in S$ with $|f(\tilde{\mathbf{x}})-f'(\tilde{\mathbf{x}})|<\varepsilon$ for all $\tilde{\mathbf{x}} \in \Omega \times X$. We have that
$$ \limsup_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} f(\Theta^n(\mathbf{x})) \leq \left(\limsup_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} f'(\Theta^n(\mathbf{x})) \right) + \varepsilon = \int_{\Omega \times X} f'(\tilde{\mathbf{x}}) \, \mu(d\tilde{\mathbf{x}}) + \varepsilon \leq \int_{\Omega \times X} f(\tilde{\mathbf{x}}) \, \mu(d\tilde{\mathbf{x}}) + 2\varepsilon $$
and likewise
$$ \liminf_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} f(\Theta^n(\mathbf{x})) \ \geq \ \int_{\Omega \times X} f(\tilde{\mathbf{x}}) \, \mu(d\tilde{\mathbf{x}}) - 2\varepsilon. $$
Since $\varepsilon$ was arbitrary, we have the desired statement.
Observation 2: For any $\omega \in \Omega$ and any probability measure $\mathbb{P}$ on $\Omega$, if there exists $y \in X$ such that $(\omega,y)$ is $(\mathbb{P} \otimes \lambda)$-generic then $(\omega,x)$ is $(\mathbb{P} \otimes \lambda)$-generic for all $x \in X$.
To see this: Fix $\omega \in \Omega$ and $x,y \in X$, and suppose that $(\omega,y)$ is $(\mathbb{P} \otimes \lambda)$-generic. Fix any $f \in C(\Omega \times X,\mathbb{R})$, and define $f' \in C(\Omega \times X,\mathbb{R})$ by $f'(\tilde{\omega},\tilde{x})=f(\tilde{\omega},\tilde{x}y^{-1}x)$. We have
$\begin{align*}
\frac{1}{N} \sum_{n=0}^{N-1} f(\Theta^n(\omega,x)) \ = \ \frac{1}{N} \sum_{n=0}^{N-1} f'(\Theta^n(\omega,y)) \
&\underset{N \to \infty}{\to} \ \int_\Omega \int_X f'(\tilde{\omega},\tilde{x}) \, \lambda(d\tilde{x}) \, \mathbb{P}(d\tilde{\omega}) \\
&= \ \int_\Omega \int_X f(\tilde{\omega},\tilde{x}) \, \lambda(d\tilde{x}) \, \mathbb{P}(d\tilde{\omega})
\end{align*}$
as required.
Completing the proof: Let $\mathbb{P}$ be a $\theta$-invariant probability measure such that $\mathbb{P} \otimes \lambda$ is $\Theta$-ergodic, and suppose for a contradiction that the conclusion of the theorem fails. By observation 1, $(\mathbb{P} \otimes \lambda)$-almost every point is $(\mathbb{P} \otimes \lambda)$-generic; and therefore, by observation 2, there exists a $\mathbb{P}$-full set $\Omega' \subset \Omega$ such that every point in $\Omega' \times X$ is $(\mathbb{P} \otimes \lambda)$-generic. Now since the conclusion of the theorem fails, there exists a $\Theta$-ergodic probability measure $\mu$ distinct from $\mathbb{P} \otimes \lambda$ such that $\mu(E \times X)=\mathbb{P}(E)$ for all $E \in \mathcal{F}$. By observation 1, there exists a $\mu$-full set $Y \subset \Omega \times X$ such that every point in $Y$ is $\mu$-generic. Obviously $Y$ is not contained in $(\Omega \setminus \Omega') \times X$ (otherwise $Y$ would be $\mu$-null), and therefore it follows that there exist points in $\Omega \times X$ that are both $(\mathbb{P} \otimes \lambda)$-generic and $\mu$-generic. However, this is not possible, since a probability measure on $\Omega \times X$ is uniquely determined by the values of the integrals that it assigns to all the members of $C(\Omega \times X,\mathbb{R})$. QED
Obviously applying the theorem to my example completes the proof of the Claim.
