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

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.