I have asked this in [MSE](https://math.stackexchange.com/questions/3016990/what-are-the-commutative-rings-r-for-which-a-otimes-bbb-z-b-a-otimes) but there was no reply. Feel free to close if inappropriate. 



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> Let $R$ be commutative ring, what can we say about the rings $R$ such that $A \otimes\_{\Bbb Z} B \cong A \otimes\_R B$ as abelian groups for all $A$ right $R$ module and $B$ left $R$ module? 
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Classify commutative rings $R$ such that $A \otimes_{\Bbb Z} B = A \otimes _{R} B$
Fernando and Pierre-Yves in the comments are right; $R$ has this property (the version where the canonical map is an isomorphism, as YCor says in the comments) iff it is a [solid ring](https://mathoverflow.net/questions/95160/solid-rings-and-tor), meaning the multiplication map $m : R \otimes\_{\mathbb{Z}} R \to R$ is an isomorphism, and no commutativity assumption is needed. To see this really clearly note that we can rewrite the canonical map as


$$\text{id}\_A \otimes m \otimes \text{id}\_B : A \otimes\_R (R \otimes\_{\mathbb{Z}} R) \otimes\_R B \to A \otimes\_R R \otimes\_R B.$$


Now the implication $\Rightarrow$ follows by letting $A = B = R$, and the implication $\Leftarrow$ follows by noting that if $m$ is an isomorphism then it's an isomorphism of $(R, R)$-bimodules. 


(The idea behind rewriting things this way is that by a variation of the [Eilenberg-Watts theorem](https://ncatlab.org/nlab/show/Eilenberg-Watts+theorem), the category of functors taking as input a left $R$-module and a right $R$-module, cocontinuous in each variable, and returning as output an abelian group is equivalent to the category of $(R, R)$-bimodules, hence any natural transformation of such functors as in the OP can be analyzed as a bimodule homomorphism, namely the bimodule obtained by substituting $A = B = R$.) 


