Let's observe what this statement is saying. Let $(\mathcal{C},\otimes)$ by a monoidal category, and $A$ be a monoid in $(\mathcal{C},\otimes)$. Let $M,N$ be objects of $(\mathcal{C},\otimes)$ with a left and right action, respectively, of $A$. Then this statement is saying that $M\otimes_{A}N$ is $M\otimes N$ quotiented by the $A$-action. But this is by construction of $M\otimes_{A}N$; see the subheading in the nlab page on the tensor product. Now if $A$ is a commutative monoid (now we're assuming $(\mathcal{C},\otimes)$ is symmetric monoidal), then $M\otimes_{A}N$ is an object of $(\mathcal{C},\otimes)$ with an action of $A$.This agrees with the monoidal structure thanks to the fact that this tensor product over $A$ can be constructed as a coequalizer $M\otimes A \otimes N \;\rightrightarrows\; M\otimes N$ given by the action of $A$ on $M$ and on $N$. Then, as the nlab says, "there is a multicategory of $A$-modules whose tensor product agrees with the coequalizer defined above".