Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$ and $$\mathcal U_{[0,\:1)}(B):=\lambda(B\cap[0,1))\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ denote the extension of the uniform distribution on $[0,1)$ to $\mathcal B(\mathbb R)$. Moreover, let $p\in[0,1]$, $\sigma>0$ and $$\kappa(x,\;\cdot\;):=p\mathcal U_{[0,\:1)}^{\otimes\mathbb N_0}+(1-p)\bigotimes_{n\in\mathbb N_0}\mathcal N_{x_n,\:\sigma^2}\;\;\;\text{for }x\in[0,1)^{\mathbb N_0}.$$ As discussed on MathOverflow, $\kappa$ is a Markov kernel with source $\left([0,1)^{\mathbb N_0},\mathcal B([0,1))^{\otimes\mathbb N_0}\right)$ and target $\left(\mathbb R^{\mathbb N_0},\mathcal B(\mathbb R)^{\otimes\mathbb N_0}\right)$.
Now let $X$ be a $[0,1)^{\mathbb N_0}$-valued random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $$\mathcal L(X)=\bigotimes_{n\in\mathbb N_0}\mathcal L(X_n)\tag0$$ and $Y$ be a $\mathbb R^{\mathbb N_0}$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname P\left[Y\in B\mid X\right]=\kappa(X,B)\;\;\;\text{almost surely for all }B\in\mathcal B(\mathbb R)^{\otimes\mathbb N_0}.\tag1$$ Are we able to show that $$\mathcal L(Y)=\bigotimes_{n\in\mathbb N_0}\mathcal L(Y_n),\tag2$$ where $\mathcal L(Z)$ denotes the distibution of a random variable $Z$ on $(\Omega,\mathcal A,\operatorname P)$, or at least that the components of $Y$ are uncorrelated?
Let $\varphi(x,\;\cdot\;)$ denote the density of $\mathcal N_{x,\:\sigma^2}$ with respect to the Lebesgue measure $\lambda$ on $\mathcal B(\mathbb R)$ for $x\in\mathbb R$. By definition, $$\kappa(x,B)=\int_B\lambda^{\otimes\mathbb N_0}({\rm d}y)\left[p\prod_{n\in\mathbb N_0}1_{[0,\:1)}(y_n)+(1-p)\prod_{n\in\mathbb N_0}\varphi(x_n,y_n)\right]\tag3$$ for all $x\in[0,1)^{\mathbb N_0}$ and $B\in\mathcal B(\mathbb R)^{\otimes N_0}$. Let $$\pi_I:\mathbb R^{\mathbb N_0}\to\mathbb R^I\;,\;\;\;x\mapsto(x_i)_{i\in I}$$ for $I\subseteq\mathbb N_0$. My idea is to show that the measures in $(2)$ coincide on $$\left\{\pi_I^{-1}\left(\prod_{i\in I}B_i\right):I\subseteq\mathbb N_0\text{ is finite and }(B_i)_{i\in I}\subseteq\mathcal B(\mathbb R)\right\}\tag4$$ from which the claim should follow. Maybe we can first apply Fubini's theorem for the product $\lambda^{\otimes\mathbb N_0}=\lambda^{\otimes I}\otimes\lambda^{\otimes\mathbb N_0\setminus I}$ and then for the product $\lambda^{\otimes I}=\bigotimes_{i\in I}\lambda$ in $(3)$ (making use of the finiteness of $I$ and that $\mathcal U_{[0,\:1)}$ and $\mathcal N_{x,\:\sigma^2}$, $x\in\mathbb R$, are both probability measures on $\mathcal B(\mathbb R)$).
By the guideline outlined above, we obtain \begin{equation}\begin{split}&\operatorname P\left[Y\in\pi_I^{-1}\left(\prod_{i\in I}B_i\right)\right]\\&\;\;\;\;\;\;\;\;\;\;\;\;=p\prod_{i\in I}\int_{B_i}\lambda({\rm d}y_i)1_{[0,\:1)}(y_i)+(1-p)\int\operatorname P\left[X\in{\rm d}x\right]\prod_{i\in I}\int_{B_i}\lambda({\rm d}y_i)\varphi(x_i,y_i)\end{split}\tag5\end{equation} and \begin{equation}\begin{split}&\left(\bigotimes_{n\in\mathbb N_0}\mathcal L(Y_n)\right)\left(\prod_{i\in I}B_i\right)\\&\;\;\;\;\;\;\;\;\;\;\;\;=\prod_{i\in I}\left(p\int_{B_i}\lambda({\rm d}y_i)1_{[0,\:1)}(y_i)+(1-p)\int\operatorname P\left[X\in{\rm d}x\right]\int_{B_i}\lambda({\rm d}y_i)\varphi(x_i,y_i)\right)\end{split}\tag6\end{equation} for all finite $I\subseteq\mathbb N_0$ and $(B_i)_{i\in I}\subseteq\mathcal B(\mathbb R)$. Now we need somehow to use $(0)$.
The claim is wrong, unless we alter the definition of $\kappa$ to $$\kappa(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N_0}\left(p\mathcal U_{[0,\:1)}+(1-p)\mathcal N_{x_n,\:\sigma^2}\right)\;\;\;\text{for }x\in[0,1)^{\mathbb N_0}.$$ With the new definition of $\kappa$ it's trivial to see that $(0)$ implies $(2)$.