Given a measurable space $(\Omega, \mathcal{F})$ with two probability measures $\mathbb{P}_1$, $\mathbb{P}_2$ and a second measurable space $(X,\mathcal{A})$ with two stochastic kernels $\mu_1, \mu_2$ from $(\Omega, \mathcal{F})$ to $(X,\mathcal{A})$, assume that the semidirect product coincides on the product space endowed with the product $\sigma$-algebra: $$\mathbb{P}_1 \times \mu_1 = \mathbb{P}_2 \times \mu_2$$ as probability measures on $(\Omega \times X, \mathcal{F} \otimes \mathcal{A})$. It then holds that $\mathbb{P}_1=\mathbb{P}_2=:\mathbb{P}$.
Question: Is it true that for $\mathbb{P}$-almost every $\omega \in \Omega$ it holds $$\mu_1(\omega, \cdot) = \mu_2(\omega, \cdot)$$ as probability measures on $(X,\mathcal{A})$ ?
So you have joint distributions $\Bbb Q_i = \Bbb P_i\otimes \mu_i$. If $\Bbb Q_1 = \Bbb Q_2$ then obviously $\Bbb P_1 = \Bbb P_2$ since they are marginals. Also, at least in cases where regular conditional distribution exists (e.g. Borel target space), you get a.s. uniqueness of it. In case you are interested only in the most general case, I am not sure whether that kind of uniqueness will hold: for sure you will get $\mu_1(\cdot, A) = \mu_2(\cdot, A)$ a.s. for all measurable $A$, but making $A$ variable may not be possible if $\Omega$ and $X$ are just some general measurable spaces.