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...
One way is to show that $M \otimes N$ modulo the the submodule $P$ generated by $m' \otimes n$ and $m \otimes n'$ (with $m' \in M', m \in M, n' \in N', n \in N$) directly satisfies the same universal property as $M/M' \otimes N/N'$.
You have a well defined $A$-bilinear map $j: M/M' \times N/N' \rightarrow (M \otimes N)/P$ given by
$$(m+M',n+N') \mapsto m \otimes n + P$$
Suppose $Q$ is an $A$-module, and $f: M/M' \times N/N' \rightarrow Q$ is $A$-bilinear. I claim that there exists a unique $A$-module homomorphism $\phi: (M\otimes N)/P \rightarrow Q$ such that $\phi \circ j = f$. The uniqueness of the tensor product will then guarantee you an isomorphism $M/M' \otimes N/N' \rightarrow (M \otimes N)/P$ such that $(m +M') \otimes (n+N') \mapsto m \otimes n + P$.
The map $M \times N \rightarrow Q, (m,n) \mapsto f(m + M',n + N')$ is certainly $A$-bilinear, so there exists a unique $A$-linear map $\psi: M \otimes N \rightarrow Q$ given on generators by $\psi(m\otimes n) = f(m+M',n+N')$. Also, for any $m' \in M', n' \in N$ you have
$$\psi(m' \otimes n) = f(m' + M',n + N') = f(0 + M', n+ N') = 0$$
$$\psi(m \otimes n') = f(m + M', n' + N') = f(m+M',0+N') = 0$$
so $P$ is contained in the kernel of $\psi$. Therefore the $A$-module homomorphism $\phi: (M \otimes N)/P \rightarrow Q$ defined by $\phi(x + P) = \psi(x)$ is well defined.
Now $\phi$ does what is required:
$$\phi \circ j(m+M',n+N') = \phi(m\otimes n + P) = \psi(m \otimes n) = j(m+M',n+N')$$
and the uniqueness of $\phi$ is easily seen from the uniqueness of $\psi$.
Best Answer
It's not true in general that tensor product commute with projective limits.
E.g. consider $\mathbb Z_p := \projlim_n \mathbb Z/p^n.$ We have that $\mathbb Z_p \otimes_{\mathbb Z} \mathbb Q$ is non-zero; it is the field $\mathbb Q_p$.
On the other hand $\mathbb Z/p^n \otimes_{\mathbb Z} \mathbb Q = 0$ for each value of $n$.
On the other hand, suppose that the modules $M_n$ are of finite length, and that $N$ is finitely presented. Then $(\varprojlim_n M_n)\otimes_A N \to \varprojlim_{n} M_n\otimes N$ is an isomorphism.
To see this, choose a finite presentation $A^r \to A^s \to N \to 0$ of $N$.
Then we have to show that the cokernel of $\varprojlim_n M_n^r \to \varprojlim_n M_n^s$ is isomorphic to the projective limit of the cokernels of the maps $M_n^r \to M_n^s$. This follows from the finite length assumption, which shows (applying Mittag--Leffler) that the projective limit of the cokernels is indeed the cokernel of the projective limits.
Now suppose that $I$ is finitely generated (e.g. assume $A$ is Noetherian). Then $A/I$ is finitely presented, and so if the $M_n$ are furthermore of finite length, the natural map you ask about is an isomorphism.