Even for Bernoulli shifts this property is not true, as the following example shows.

Let $(X,\mu)$ be the $(\frac 12,\frac 12)$ Bernoulli shift on two symbols, 0 and 1. That is, $X = \{0,1\}^\mathbb{N}$ is the set of all one-sided infinite sequences of 0s and 1s, and $\mu$ gives weight $2^{-n}$ to every $n$-cylinder.

Let $A=[1]$ be the set of sequences beginning with the symbol 1. Interpreting $x\in X$ as a series of coin flips, with 0 as heads and 1 as tails, $A$ corresponds to the event that the first flip is tails.

Let $B$ be the set of sequences $x\in X$ with the property that for every $k=1,2,3,\dots$, there is some index $i\in [2^{k-1}, 2^k) \cap \mathbb{N}$ such that $x_i = 1$. Then $B$ corresponds to the event that the first flip is tails, then (at least) one of the next two flips is tails, then (at least) one of the next four is tails, and so on.

Note that $[0] \cap B = \emptyset$ and so $\mu(B) \leq \frac 12$. On the other hand, $X \setminus B \subset [0] \cup \sigma^{-1}[00] \cup \sigma^{-3}[0000] \cup \cdots$, and so $\mu(X\setminus B) \leq \sum_{k=0}^\infty 2^{-2^k} < 1$. Thus $0 < \mu(B) < 1$.

Now consider $B \cap \sigma^{-n}(A)$. This is the event that $B$ holds (first flip tails, one of next two tails, etc.), and also that the $n$th flip is tails. Let $k'$ be the unique integer with $n\in [2^{k'},2^{k'+1})$. Considering conditional probabilities, we have
$$
\frac{\mu(B\cap \sigma^{-n}(A))}{\mu(A)} =
\mathbb{P}(B \mid \sigma^{-n}(A)) = \prod_{k\neq k'} (1-2^{-2^k})
> \prod_{k} (1-2^{-2^k}) = \mathbb{P}(B) = \mu(B).
$$
In particular,
$$
\mu(B\cap \sigma^{-n}(A)) > \mu(A) \mu(B),
$$
and this holds for every $n$.

There may be simpler examples. I don't know how to generalise this to arbitrary measure-preserving transformations, but my expectation would be that for every mpt there are measurable sets $A,B$ such that you do not get equality in the mixing condition for any finite $n$.