The answer is ‘yes, it is necessary’.
Hint:
Take $J=0$. The assertion would mean that $\operatorname{Supp}M=V(\operatorname{Ann}_AM)$, even if $M$ is not finitely generated.
However, in Bourbaki, Commutative Algebra, ch. II, Localisation, §4, exercise 22 c), you have the following counter-example: let $p$ be a prime number and set
$$M=\bigoplus_{k\in\mathbf N}\mathbf Z/p^k\mathbf Z.$$
Then $\operatorname{Supp}M$ is closed, but distinct from $V(\operatorname{Ann}_{\mathbf Z}M)$.
First, the universal property says that for any homomorphism $f: A\to B$ such that $f(S)\subset B^*$, the grup of units of $B$, there is a unique homomorphism $f': A[S^{-1}]\to B$ such that $f=f'\circ j_S$.
Now, on one hand we have a homomorphism $jj: A[S^{-1}]\to A[S^{-1}][j_S(T)^{-1}]$ and the composition $h=jj\circ j_S$ send clearly the elements of $T$ to invertible elements. So by the universal property, there exist a unique homomorphism say $h': A[T^{-1}] \to A[S^{-1}][j_S(T)^{-1}]$ such that $h=h'\circ j_T$.
On the other hand, the homomorphism $j_T$ sends the elements of $S$ to invertible elements, so we have a map $j'_T: A[S^{-1}]\to A[T^{-1}]$ such that $j_T=j_T'\circ j_S$. This homomorphism sends the elements of $j_S(T)$ to invertible elements, so we have another homomorphism $ i: A[S^{-1}][j_S(T)^{-1}] \to A[T^{-1}]$ such that $j_T'=i\circ jj$.
Now, one only needs to show that $i\circ h'$ is the identity, as well as $h'\circ i$. But this is deduced from the unicity of the universal property.
For the first composition: from $h=jj\circ j_S$, $h=h'\circ j_T$, $j_T'=i\circ jj$ and $j_T=j_T'\circ j_S$, we have $$j_T=(i\circ jj) \circ j_S=i\circ (jj \circ j_S)=i\circ h=(i\circ h')\circ j_T.$$
So $i\circ h':A[T^{-1}]\to A[T^{-1}]$ is a map that verifies the property of $(j_T)'$ corresponding to $j_T$ (it is different from $j'_T$, sory for the messing notation), which also the map $\operatorname{id}$ verifies. So by unicity it must be the identity.
For the second composition $$h'\circ i: A[S^{-1}][j_S(T)^{-1}] \to A[S^{-1}][j_S(T)^{-1}] $$
note first that $jj=h'\circ j'_T$, by its universal property, since
$$h'\circ j'_T\circ j_S=h'\circ j_T=h=jj\circ j_S$$ and hence it verifies the same property that $jj$, which is uniquely determined by this property. But then
we have $$jj=h'\circ j'_T=h'\circ (i\circ jj)=(h'\circ i) \circ jj$$
and we can argue as the first composition.
Best Answer
Going off user26857's comment, we provide a counterexample for Proposition 3.14 in the case that $M$ is not finitely generated. Hopefully you can use this to construct a counterexample for Corollary 3.15 as well.
Take $A = \mathbb{Z}$, and let $M$ be the direct sum of $\mathbb{Z}/k\mathbb{Z}$ as $k$ ranges through $\mathbb{N}$. This is not finitely generated as a $\mathbb{Z}$-module (there are infinitely many nonzero summands). Now, what is the annihilator of this module? Well, it is just $0$, since for any integer $n$ different from $0$, $n$ does not act by $0$ on the $\mathbb{Z}/(n+1)\mathbb{Z}$ factor, so $n$ does not act by $0$ on $M$. (If $n$ is negative, look at $\mathbb{Z}/(1-n)\mathbb{Z}$. Strictly, this is not necessary since any ideal of $\mathbb{Z}$ is generated by a positive number, but oh well.)
Now, let us localize at $S = \mathbb{Z} \setminus \{0\}$, so $S^{-1}A = \mathbb{Q}$ and $S^{-1}\text{Ann}(M) = S^{-1}0 = 0$.
But what is $S^{-1}M$? We claim that it is $0$. This is not hard to show. Let $e_k$ be the generator of the copy of $\mathbb{Z}/k\mathbb{Z}$ in $M$. Then the set $\{e_k\}$ is a $\mathbb{Z}$-generating set for $M$ in the sense that every element of $M$ is a finite linear combination of the elements $e_k$ with coefficients in $\mathbb{Z}$, so $\{e_k/1\}$ is a $\mathbb{Q}$-generating set for $S^{-1}M$ (in this same sense).
But for each $k$, we have $k \cdot e_k = 0$ in $M$, which shows that $e_k/1 = 0$ in $S^{-1}M$. To be a little more explicit, we want to show that $e_k/1 = 0/1$ in $S^{-1}M$. By definition, this happens if and only if there is some $s \in S$ such that $s(1 \cdot e_k - 0 \cdot 1) = 0$. Now take $s = k$.
So now, we have that $\{0\}$ is a $\mathbb{Q}$-generating set for $S^{-1}M$. Thus, $S^{-1}M = 0$, so $\text{Ann}(S^{-1}M) = \mathbb{Q}$, which is different from $S^{-1}\text{Ann}(M) = 0$. (Here, we need the crucial fact that there exist nonzero rational numbers, i.e., that $\mathbb{Q}$ is not equal to $0$.)