The answer is NOT NECESSARY (at least for $3\times 3$ matrices).

Let $A=\left(\matrix{2&0&1\\\0&1&0\\\0&0&\alpha}\right)$, $B=\left(\matrix{0&1&1\\\1&0&0\\\0&0&\beta}\right)$, $v=\left(\matrix{1\\\1\\\0}\right)$ (here $\alpha$, $\beta$ are algebraically independent over $\mathbb Q$). Then $A^nBA^nv=2^nv$, so $A^nBA^n\in E$. Next, our monoid is freely generated by $A$ and $B$. So, the element $A^nBA^n$ can be expanded as a non-trivial product of two elements in $E$ iff it can be expanded literally; this is obviously false.

So, $E$ is not finitely generated in this case.

UPDATE: Let us show that the monoid $(A,B)$ is free. Actually, we will show that every word $W=W(A,B)$ can be recovered from its action on $u=\left(\matrix{0\\\0\\\1}\right)$. Let $Wu=u'=\left(\matrix{u_1'\\\u_2'\\\u_3'}\right)$; since $A$, $B$ are invertible, it is enough to reconstruct the first letter in $W$.

So let $W=XW'$, $X$ is the first letter. Note that $u_3'=\alpha^k\beta^\ell$, where $k$ and $\ell$ are the numbers of occurences of $A$ and $B$ in $W$; hence we have reconstructed $k+\ell$. Next, on each of the previous steps, to the first two coordinates we added only the monomials in $\alpha$ and $\beta$ of lower degrees; so, $u_1'$ and $u_2'$ are polynomials in $\alpha$ and $\beta$ of degree $\leq k+\ell-1$. Moreover, the leading term in $u_1$ is either $\alpha^k\beta^{\ell-1}$ or $\alpha^{k-1}\beta^\ell$, depending on the number of letters $A$ and $B$ in $W'$, that is --- depending on $X$. Hence $X$ is reconstructible.

Surely, we used several times that $\alpha$, $\beta$ are algebraically independent over $\mathbb{Q}$. In fact, by the same lines one may prove the same for $\alpha=1/2$ and $\beta=1/3$ ;)...