What are the commutative rings $R$ for which $A \otimes _{\Bbb Z} B = A \otimes _R B$ as abelian groups? This is a follow up. 

What are the commutative rings $R$, for which given $R$-modules $A$ and $B$,  $A \otimes _{\Bbb Z} B = A \otimes _R B$ as abelian groups? 


This is true when $R= \Bbb Q$, or $\Bbb Z_m$. We can give an $R$-module structure $A \otimes _{\Bbb Z} B$ satisfying $r (a \otimes b) = ra \otimes b$. 
When $R= \Bbb Q$, we get the additional fact that $a \otimes rb=ra \otimes b$. To see this,  note that when $r \in \Bbb N$, we have 
$$ r a \otimes b = \sum a \otimes b = a \otimes rb $$ by bilinearity  - which extends $r$ to $\Bbb Z$ too. When $r=1/m$, $m \in \Bbb Z$, 
$$ \frac{1}{m}a \otimes b = \frac{1}{m} (a \otimes b) = \frac{1}{m} ( \sum (a \otimes \frac{1}{m} b)) = \frac{1}{m} (ma \otimes \frac{1}{m} b ) = a \otimes \frac{1}{m} b $$ 
Thus, we have equality for all $r \in \Bbb Q$. 

I think generalizing to $\Bbb Q$ is as far as we can get for this naive strategy. I wonder if there exists better method for the classification. 
 A: First, there is an ambiguity in your question. The usage of $=$ is ambiguous, but I'll interpret it as meaning that the natural map $\newcommand\tens\otimes\newcommand\ZZ{\mathbb{Z}}A\tens_\ZZ B\to A\tens_R B$ is an isomorphism. 
Given this interpretation, we can give a simple criterion which rings with this property must satisfy. Namely the natural map $R\tens_\ZZ R\to R$ must be an isomorphism. This is also sufficient though, since if this is true, then
$$A\tens_\ZZ B \simeq (A\tens_R R)\tens_\ZZ (R\tens_R B)\simeq A\tens_R (R\tens_\ZZ R)\tens_R B \simeq A\tens_R R \tens_R B \simeq A\tens_R B,$$
where I'm using $\simeq$ for natural isomorphism.
Thus the question is reduced to the question of for which rings is the natural map $R\tens_\ZZ R \to R$ an isomorphism. This is equivalent to asking for which rings is the diagram
$$\newcommand\id{\operatorname{id}}
\require{AMScd}
\begin{CD}
\ZZ @>\iota>> R \\
@V\iota VV @VV\id V \\
R @>\id >> R
\end{CD}
$$
a pushout diagram (where $\iota$ is the unique map $\ZZ\to R$.
Well, if it is, then for any pair of morphisms $f,g : R\to S$ with $f\circ \iota = g\circ \iota$, then there is a unique map $h : R\to S$ with $f=h\circ \id = g$. Thus $\iota$ is an epimorphism.
Conversely if $\iota$ is an epimorphism, then for any pair of morphisms $f,g : R\to S$ with $f\circ \iota = g\circ \iota$, then $f=g$, so the map $h=f=g : R\to S$ satisfies $f=h\circ\id$ and $g=h\circ\id$. Thus if $\iota$ is epic, this diagram is a pushout.
Hence a commutative ring $R$ has the property that $A\tens_\ZZ B\simeq A\tens_R B$ for all pairs $A$ and $B$ of $R$-modules if and only if the natural map $\ZZ\to R$ is an epimorphism.
Edit
As for what rings $R$ for which the natural map $\ZZ\to R$ is an epimorphism look like, I'm not sure. In general epis in the category of rings are complicated. That said, if I had to guess the answer in this case, my guess would be that these rings would be the subrings of $\Bbb{Q}$ and the rings $\ZZ/n\ZZ$, but that should probably be another question.
