Let $\varrho:A\to B$ an homomorphism of commutative rings. For each $A$-module $M$ let $\eta_M:M\to B\otimes_AM$ denote the canonical $A$-module homomorphism $x\mapsto 1\otimes x$.
It's know that when $M$ is a $B$-module, hence an $A$-module by scalar restriction trough $\varrho$, then $\eta_M$ is a split monomorphism of $A$-module with retraction given by \begin{align} &B\otimes_AM\to M& &b\otimes x\mapsto bx \end{align} If, moreover, $\varrho$ is a ring localization, then $\eta_M$ is a $B$-module isomorphism.
I ask if this occurs whenever $\varrho:A\to B$ is a ring epimorphism, thus my question is:
Let $\varrho:A\to B$ be an epimorphism of commutative rings and $M$ be a $B$-module. The canonical $A$-module homomorphism $\eta_M:M\to B\otimes_A M$ is a $B$-module isomorphism?
My try. It's know that $\eta_B:B\to B\otimes_AB$ is an isomorphism of $A$-algebras because $b\otimes 1=1\otimes b$ in $B\otimes_AB$ for all $b\in B$. Tensoring with $B$ the canonical $B$-module homomorphism $\bigoplus_MB\twoheadrightarrow M$ we get \begin{align} \bigoplus_MB &\cong\bigoplus_A(B\otimes_AB)\\ &\cong B\otimes_A\bigoplus_MB \end{align} hence the commutative diagram below shows that $\eta_M:M\to B\otimes_AM$ is surjective, hence bijective.
As far as I can tell your proof is correct. Another approach was hinted at in the comments by user119882: By using that $B \cong B \otimes_A B$ as $B$-algebras we have the isomorphisms of $B$-modules $$ B \otimes_A M \cong B \otimes_A B \otimes_B M \cong B \otimes_B M \cong M $$ which is given by simple tensors by $$ b \otimes m \mapsto b \otimes 1 \otimes m \mapsto b \otimes m \mapsto bm. $$
I would also like to add an argumant as to why the homomorphism of $B$-algebras $$ \Phi \colon B \otimes_A B \to B, \quad b_1 \otimes b_2 \mapsto b_1 b_2 $$ is already an isomorphism: The two maps $f, g \colon B \to B \otimes_A B$ given by $$ f(b) = b \otimes 1 \quad\text{and}\quad g(b) = 1 \otimes b $$ are ring homomorphisms with $f \circ \varrho = g \circ \varrho$. Because $\varrho$ is an epimorphism, it follows that $f = g$, and thus $$ 1 \otimes b = b \otimes 1 $$ for all $b \in B$. It then follows for all $x, y, b \in B$ that $$ (xb) \otimes y = (x \otimes y)(b \otimes 1) = (x \otimes y)(1 \otimes b) = x \otimes (yb). $$ The map $$ \Psi \colon B \to B \otimes_A B, \quad b \mapsto b \otimes 1 $$ is therefore an inverse of $\Phi$, since $$ \Phi(\Psi(b)) = \Phi(b \otimes 1) = b $$ and $$ \Psi(\Phi(b_1 \otimes b_2)) = \Psi(b_1 b_2) = (b_1 b_2) \otimes 1 = b_1 \otimes b_2. $$