I think the martingale theory really captures exactly what's happening. I don't know any other proof, but suspect that you would end up reproducing the backwards martingale theorem in some form? I am reproducing the proof below, which I think is elementary other than one use of the backwards martingale theorem, and assuming some properties of conditional expectation.
As Ronnie says, your condition implies that the tail $\sigma$-algebra, $\bigcap_n T^{-n}\mathcal B$ is trivial: if $A\in \bigcap_{n\ge 0} T^{-n}\mathcal B$ is a set of positive measure, then for each $n$, $A=T^{-n}B_n$ for some set $B_n$, and $T^nA=B_n$, so that by your assumption $\mu(B_n)\to 1$. Since $\mu(B_n)=\mu(A)$, it follows that $\mu(A)=1$.
Now $T^{-n}\mathcal B$ is a decreasing sequence of $\sigma$-algebras whose limit is the trivial $\sigma$-algebra. It follows that for any $f\in L^2$, $\mathbb E(f|T^{-n}\mathcal B)$ converges in $L^2$ to $\int f\,d\mu$ by the backwards martingale theorem. Taking $f=\mathbf 1_A$ and setting $g=\mathbf 1_B$, we have
$$
\mu(A\cap T^{-n}B)=\mathbb E(g\circ T^nf)=\mathbb E(\mathbb E(g\circ T^nf|T^{-n}\mathcal B))=\mathbb E(g\circ T^n\cdot\mathbb E(f|T^{-n}\mathcal B)).
$$
We used the "tower law" for the second equality and brought out a $T^{-n}\mathcal B$-measurable "constant" in the third equality.
Since $\mathbb E(f|T^{-n}B)$ converges in $L^2$ to $\mu(A)$, it is easy to check that the final expectation converges to $\mu(A)\mathbb E(g\circ T^n)=\mu(A)\mu(B)$.
