Note: $A$ and $B$ are assumed to be commutative, if this matters. Moreover, I'm fairly certain flatness of $B$ is irrelevant to this, but this question is part of a larger problem with this hypothesis so I included it just in case.
Let $x\in M$ be nonzero. Let $M_B$ be the $B$-module $M\otimes_A B$, and so $1\otimes x\in M_B$. Take the map $A\to Ax$ defined in the obvious way. If this has kernel $\mathfrak a$, then $A/\mathfrak a\cong Ax$. In this way it's clear that $\mathfrak a^e$ is contained in the kernel of $B\to B(1\otimes x)$. Explicitly, if $$b = \sum_{\substack{a_i\in\mathfrak a\\ b_i\in B}}f(a_i)b_i,$$ then $$b(1\otimes x)= \sum_{\substack{a_i\in\mathfrak a\\ b_i\in B}}(b_i\otimes a_ix) = 0.$$
However, I cannot see why the kernel is precisely $\mathfrak a^e$. I can buy that $1\otimes y = 0$ if and only if $y = 0$. Hence if $b = f(a)$, then $0=b\otimes x = 1\otimes ax$ implies that $ax = 0$ and $a\in\mathfrak a$. But why couldn't there be, for example, some $b$ not in the image of $f$ such that $b\otimes x = 0$? I don't see why $b\otimes x$ implies that $b$ is some linear combination of the $f(a_i)$ for $a_i\in\mathfrak a$.
There is an exact sequence $$0\to\mathfrak{a}\to A\stackrel{x}\to M.$$ Since $B$ is flat, this exact sequence remains exact after tensoring with $B$, giving an exact sequence $$0\to\mathfrak{a}\otimes B\to B\stackrel{1\otimes x}\to M_B.$$ The image of the map $\mathfrak{a}\otimes B\to B$ is $\mathfrak{a}^e$, so this says the kernel of $B\to B(1\otimes x)$ is $\mathfrak{a}^e$.
Note that flatness of $B$ is crucial here. If $M$ was generated by $x$, then flatness of $B$ would not be necessary, since our original exact sequence would be short exact, and so would remain exact in the middle (though not necessarily on the left) after tensoring with $B$, which is all we need for our conclusion. But since $M$ is not necessarily generated by $x$, you need flatness of $B$ to get even the exactness in the middle. For a counterexample, for instance, take $A=\mathbb{Z}$, $B=\mathbb{Z}/(2)$, $M=\mathbb{Z}$, and $x=2$.