If $a_M$ is not locally nilpotent, why does there exist $x \in M$ and a prime ideal $\mathfrak{p}$ such that $(Ax)_\mathfrak{p} \neq 0$?

I think I understand this now. It follows from the definitions that if $x$ is an element of $M$ with $\frak{a}$ its annihilator, and $p$ a prime ideal of $A$, then $(Ax)_p\neq 0$ iff $p\supset\frak{a}$.

Now $S$ is disjoint from the annihilator of $x$, so the annihilator of $x$ is contained in some ideal maximal among those not intersecting $S$, which happens to be prime. Take $p$ to be this prime ideal, and so $(Ax)_p\neq 0$.

For any $A$-module $M$ let $S(M)$ be its support and $\mathfrak a(M)$ its annihilator. Let $LN$ be the set of those $a$ in $A$ such that $a_M$ is locally nilpotent. We want to prove $$ LN=\bigcap_{\mathfrak p\in S(M)}\ \mathfrak p.\qquad\qquad(1) $$ The main lemma will be:

$(2)$ If $M$ is finitely generated, then $S(M)=V(\mathfrak a)$, with $\mathfrak a:=\mathfrak a(M)$.

Recall: $V(\mathfrak a):=\{\mathfrak p\in\text{Spec}(A)\ |\ \mathfrak p\supset\mathfrak a\}$.

Proof of $(2)$. Assume $\mathfrak p\supset\mathfrak a$. We must show $M_{\mathfrak p}\neq0$. Suppose by contradiction $M_{\mathfrak p}=0$. Let $x_1,\dots,x_n$ be generators of $M$. For each $i=1,\dots,n$ there is an $s_i$ which is $A$ but not in $\mathfrak p$ such that $s_ix_i=0$. Then $s_1\cdots s_n$ is in $\mathfrak a$ but not in $\mathfrak p$, contradiction.

Conversely, assume $M_{\mathfrak p}\neq0$. Then there is an $x$ in $M$ such that $sx\neq0$ for all those $s$ in $A$ which are not in $\mathfrak p$. Let $\alpha\in\mathfrak a$. As $\alpha x=0$, the element $\alpha$ is in $\mathfrak p$.

Proof of $(1)$. Let $(M_i)$ be a family of finitely generated submodules of $M$ whose sum is $M$. Put $\mathfrak a_i:=\mathfrak a(M_i)$.

By definition we have $$ LN=\bigcap_i\ r(\mathfrak a_i). $$ Recall the equality $$ r(\mathfrak a_i)=\bigcap_{\mathfrak p\in V(\mathfrak a_i)}\ \mathfrak p. $$ where $r(\mathfrak a_i)$ is the radical of $\mathfrak a_i$. By $(2)$, we have $$ \bigcap_{\mathfrak p\in V(\mathfrak a_i)}\ \mathfrak p=\bigcap_{\mathfrak p\in S(M_i)}\ \mathfrak p. $$ For formal reasons we have $$ \bigcap_i\ \bigcap_{\mathfrak p\in S(M_i)}\ \mathfrak p=\bigcap_{\mathfrak p\in\bigcup S(M_i)}\ \mathfrak p. $$ Now $(1)$ follows from the fact that $S(M)$ is the union of the $S(M_i)$.