It is not generally true that $\mathcal{A}_n\times\mathcal{B}$ decreases to $\mathcal{A}_\infty\times\mathcal{B}$. In general, the inequality $\mathcal{A}_\infty\times\mathcal{B}\subseteq\bigcap_n(\mathcal{A}_n\times\mathcal{B})$ holds, but the inequality can be strict. I'll demonstrate this by a counterexample.
Consider the case where $X=Y=2^\mathbb{N}$. An element $\omega$ of this space is a function $\omega\colon\mathbb{N}\to\{0,1\}$. Think of this as the space of an infinite sequence of coin tosses, where $\omega(n)=1$ if the n'th toss is a head and $\omega(n)=0$ if it is a tail. Also, let $\mathcal{A}=\mathcal{B}$ be the sigma-algebra generated by the individual tosses, so that it is generated by the sets $A_n=\{\omega\in X\colon\omega(n)=1\}$ ($n=1,2,\ldots$). Also, let $\mathcal{A}_n$ be the sigma-algebra generated by all the tosses on or after the n'th one. So, $\mathcal{A}_n$ is generated by $\{A_m\colon m\ge n\}$. Then, $\mathcal{A}_\infty=\bigcap_n\mathcal{A}_n$ is the tail sigma-algebra. I claim that the set $ S=\bigcup_{n=1}^\infty\left\{(\omega_X,\omega_Y)\in X\times Y\colon \omega_X(m)=\omega_Y(m)\;{\rm for\ all\;}m\ge n\right\} $ is in $\bigcap_n(\mathcal{A}_n\times\mathcal{B})$ but not in $\mathcal{A}_\infty\times\mathcal{B}$. Here, $S$ represents the pairs of sequences of coin tosses which agree with each other for all but finitely many tosses. The event for which they agree on every toss starting from the n'th one is $S_n=\bigcap_{m\ge n}((A_m\times A_m)\cup(A_m^c\times A_m^c))$, which is in $\mathcal{A}_n\times\mathcal{B}$. We then have $S=\bigcup_{m\ge n}S_m\in\mathcal{A}_n\times\mathcal{B}$ for all $n$, so $S\in\bigcap_n(\mathcal{A}_n\times\mathcal{B})$. In fact, $S\in\bigcap_n(\mathcal{A}_n\times\mathcal{A}_n)$.
We can also show that $S\not\in\mathcal{A}_\infty\times\mathcal{B}$. Let $\mathbb{P}$ be the probability measure on $(X\times Y,\mathcal{A}\times\mathcal{B})$ such that, the outcomes $(\omega_X,\omega_Y)\in X\times Y$ are distributed such that $\omega_X(n)$ are independent Bernoulli random variables with probability 1/2 of being equal to one (representing tossing a fair coin), $\omega_Y=\omega_X$ with probability 1/2, and $\omega_Y=1-\omega_X$ with probability 1/2. That is, if $\epsilon=(\epsilon_1,\epsilon_2,\ldots)$ is a sequence of independent random variables with $\mathbb{P}(\epsilon_k=1)=\mathbb{P}(\epsilon_k=0)=1/2$ then $ \mathbb{P}(U)=\frac12\mathbb{P}\left((\epsilon,\epsilon)\in U\right)+\frac12\mathbb{P}\left((\epsilon,1-\epsilon)\in U\right) $ for any $U\in\mathcal{A}\times\mathcal{B}$. Then, $\mathbb{P}(S)=1/2$. Now consider any set $U=V\times W\in\mathcal{A}_\infty\times\mathcal{B}$. Kolmogorov's zero-one law says that $V$ has probability 0 or 1. So, up to a set of zero probability, we have $U=X\times W$ or $U=\emptyset\times W=X\times\emptyset$. In any case, up to set of zero probability, $U$ can be written as $X\times W$ for some $W\in \mathcal{B}$ and, by the monotone class theorem (or pi-system d-system lemma), this extends to all $U\in\mathcal{A}_\infty\times\mathcal{B}$. However, $S$ is not of this form. In fact, as $S$ only depends on the events $C_n=\{(\omega_X,\omega_Y)\colon\omega_X(n)=\omega_Y(n)\}$, it can be seen that it is independent of $\{\emptyset,X\}\times\mathcal{B}$ so, if it did lie in this sigma-algebra (up to a probability zero event) then it would be independent of itself. This would imply $\mathbb{P}(S)=0$ or $\mathbb{P}(S)=1$, which is a contradiction.
It is true that, if $\mu,\nu$ are probability measures on $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ respectively and $\mathbb{P}=\mu\times\nu$ is the product measure, then every $S\in\bigcap_n(\mathcal{A}_n\times\mathcal{B})$ is in $\mathcal{A}_\infty\times\mathcal{B}$ up to a zero probability set. In fact, you can show that $ \mathbb{E}\left[X\;\big\vert\mathcal{A}_n\times\mathcal{B}\right]\to\mathbb{E}\left[X\;\big\vert\mathcal{A}_\infty\times\mathcal{B}\right] $ is probability as $n\to\infty$, for any integrable $\mathcal{A}\times\mathcal{B}$-measurable random variable $X$ (in fact, it converges almost-surely, but that is not needed here). In particular, if $X$ is $\bigcap_n(\mathcal{A}_n\times\mathcal{B})$-measurable, then the left hand side is just $X$, so $X=\mathbb{E}[X\;\vert\mathcal{A}_\infty\times\mathcal{B}]$ is (almost-surely) $\mathcal{A}_\infty\times\mathcal{B}$-measurable. To prove this limit, it is enough to consider $X=YZ$ where $Y$ $\mathcal{A}$-measurable and $Z$ is $\mathcal{B}$-measurable, and apply the monotone class lemma. You will probably need Levy's downward theorem to show that $\mathbb{E}[Y\;\vert\mathcal{A}_n]$ tends to $\mathbb{E}[Y\;\vert\mathcal{A}_\infty]$ in probability (I have a proof of the downward theorem on by blog, here).