This is an excellent question, and your intuition is not too far off.
Note that the counterexamples given in the other answers involve a non-finitely generated module ($\mathbb Q$ thought of as a $\mathbb Z$-module), and constructions like annihilators behave much better for finitely generated modules than for non-finitely generated ones (as a general rule).
E.g. Suppose that $M = R/I$ and $N = R/J$ are cyclic. Then $M\otimes N = R/(I+J),$ and so your conjectured formula is true in that case.
As you observe, always $\operatorname{Ann} M + \operatorname{Ann} N \subset
\operatorname{Ann} M\otimes N$. Regarding the converse,
what is true for general finitely generated $R$-modules $M$ and $N$ is that $\operatorname{Ann}M\otimes N$ is contained in the radical of $\operatorname{Ann} M + \operatorname{Ann} N$. So your conjecture becomes correct if you restrict to f.g. modules, and take radicals of both sides.
[To see this, use the fact that for a finitely generated module, the radical of $\operatorname{Ann} M$ is equal to the intersection of those prime ideals $\mathfrak p$ for which $\kappa(\mathfrak p) \otimes_R M \neq 0$, where $\kappa(\mathfrak p)$ denotes the fraction field of $R/\mathfrak p$.]
I don't think that we can do much better than this though, in the finitely generated but non-cyclic case.
To see the kind of things that can happen, choose three ideals $I_1,I_2,J,$
and let $M = R/I_1 \oplus R/I_2$ and $N = R/J$. Then $\operatorname{Ann} M = I_1\cap I_2$, $\operatorname{Ann} N = J$, and $M\otimes N = R/I_1+J \oplus R/I_2 + J$, so that $\operatorname{Ann} M\otimes N = (I_1 + J) \cap (I_2 +J).$
Now we always have $(I_1\cap I_2) + J \subset (I_1 + J) \cap (I_2 + J)
\subset \operatorname{rad}\bigl((I_1 \cap I_2) + J\bigr),$ but the first inclusion is typically not an equality (because the lattice of ideals in $R$ is always modular, but typically not distributive).
So if you take a counterexample to the distributive property for some lattice of ideals, feeding it into the above example will give a counterexample to the precise form of your conjecture (i.e. equality on the nose, rather than up to taking radicals) involving only finitely generated modules.
Given that $M$ and $N$ are finitely generated, we have that $M = A \langle x_1, \dots, x_m \rangle$ for some elements $x_i$ in $M$ and $N = A \langle y_1, \dots, y_n \rangle$ for some elements $y_j$ in $N.$ Consider the surjections $\pi : A^m \to M$ and $\rho : A^n \to N$ defined by $\pi(r_1, \dots, r_m) = r_1 x_1 + \cdots + r_m x_m$ and $\rho(r_1, \dots, r_n) = r_1 y_1 + \cdots + r_n y_n.$ Recall that the functors $M \otimes_A -$ and $- \otimes_A N$ are right-exact, hence we have surjections $\pi \otimes_A 1_N : A^m \otimes_A N \to M \otimes_A N$ and $1_M \otimes_A \rho : M \otimes_A A^n \to M \otimes_A N.$ By hypothesis that there exists an isomorphism $\varphi : M \otimes_A N \to A,$ it follows that $\varphi \circ (\pi \otimes_A 1_N) : A^m \otimes_A N \to A$ and $\varphi \circ (1_M \otimes_A \rho) : M \otimes_A A^n \to A$ are surjections. Certainly, $A$ is a free $A$-module, hence the maps $\varphi \circ (\pi \otimes_A 1_N)$ and $\varphi \circ (1_M \otimes_A \rho)$ split, i.e., there exist $A$-modules $M'$ and $N'$ such that $M \otimes_A A^n \cong M' \oplus A$ and $A^n \otimes_A M \cong A \oplus N'.$ Putting this all together gives $$A^n \cong (M \otimes_A N) \otimes_A A^n \cong N \otimes_A (M \otimes_A A^n) \cong N \otimes_A (M' \oplus A) \cong N \oplus (N \otimes_A M'),$$ and analogously, we have that $A^m \cong M \oplus (M \otimes_A N').$ Consequently, both $M$ and $N$ are direct summands of a free $A$-module, so they are projective $A$-modules. But a finitely generated projective module over a Noetherian local ring is free, so we are done.
Best Answer
No, we may not in general make the identification evoked in your more general question.
For example take $A=\mathbb R , B=\mathbb C , M=N =\mathbb C$. Then $\mathbb C \otimes_{\mathbb C} \mathbb C =\mathbb C$ whereas $\mathbb C \otimes_{\mathbb R} \mathbb C=\mathbb C \times \mathbb C $.
But sometimes we may...