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No, that is not true. Let $A$ be $\mathbb{Z}[x]/\langle x(1-x) \rangle$. Let $B$ be $A/\langle x \rangle$ with the obvious quotient morphism, $\Phi$. Let $f$ be $x$. Then the natural $A$-algebra homomorphism, $$ A[y]/\langle yx-1 \rangle \to A/\langle 1-x \rangle, \ \ y \mapsto 1, $$ is an isomorphism. To see this, observe that in $A[y]/\langle yx-1 \rangle$ we have the congruences, $$ 1-x = 1(1-x) \equiv (yx)(1-x) = y(x(1-x)) = y\cdot 0 = 0.$$ So let $M$ be $A[y]/\langle xy-1 \rangle = A/\langle 1-x \rangle$. Then $M\otimes_A B$ equals $A/\langle 1-x \rangle \otimes_A A/\langle x \rangle$, which is $\{0\}$. But $M$ is not zero.