It is easy to see that it is true for commutative rings under the canonical map $A\otimes s\mapsto sA$. When I wanted to generalize to noncommutative ring. Then I see this map is not really multiplicative: $(A\otimes x)(B\otimes y)\mapsto xAyB$ but not really equal to the image of $AB\otimes xy$ which is $xyAB$.
Eventually, I figured out why this is a bit "weird". Let's say $S$ is a noncommutative $R$-algebra. We need to be careful with tensor products. So usually, it is fine that we can just use the rule that $ra\otimes b=a\otimes rb$. When $R,S$ are non-commutative, we should use this rule instead $ra\otimes b=a\otimes br$.
Even this, we still get some problems when considering extension of scalars. Extension of scalar is totally fine if we only consider $S$ as a module. There is some problem for the multiplication.
Take $R=\mathbb C$ and $S=\mathbb H$ the quaternion algebra. We embed $R\hookrightarrow S$ by sending $i$ to $I$. If we consider $\mu:\mathbb C\otimes_{\mathbb C}\mathbb H\to \mathbb H, a\otimes b\mapsto ab$ (or also $ba$). Then there is a problem. Consider $1\otimes j$ and $i\otimes 1$ $(1\otimes j)(i\otimes 1)=(i\otimes j)\mapsto ij$. However, $\mu(1\otimes j)\mu(i\otimes 1)=ji$.
So is there any way to fix this?