Canonical homomorphism into scalar extension module through a ring epimorphism 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.

 A: 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
$$
    \Psi(\Phi(b_1 \otimes b_2))
  = \Psi(b_1 b_2)
  = (b_1 b_2) \otimes 1
  = b_1 \otimes b_2.
$$
