Yes, you can recover the $A$-module $V$. There is a minimal ideal in $End(V)$, namely the ideal $F$ of finite-rank-operators. It is in fact the *unique* minimal two-sided ideal, because for every $f\neq 0$ one can find a finite-rank operator $g$ with $gf\neq 0$.

$F$ is isomorphic to the algebraic tensor product $V \otimes_{alg} V^\ast$ as a $A$-$A$-bimodule via the canonical map $v\otimes\phi \mapsto \phi(-)v$. This also proves that $A$ acts transitively on $V\setminus\{0\}$. One can readily verify that the left ideals contained in $F$ are in 1-to-1 correspondence to finite-dimensional quotients of $V$ via $(V\xrightarrow{q}\mathbb{C}^n) \leftrightarrow \{f \mid f\text{ factors through }q\}$ and this left ideal is isomorphic to $\sum_{i=1}^n V\otimes q_i \leq V\otimes V^\ast$ as $A$-modules. In particular, all the minimal left ideals contained in $F$ are isomorphic to $V$ as $A$-modules.

Now if you want $End(V)$ to also recover the topology of $V$, you need more information, I think. Of course you can recover $V^\ast$ as a vector space and the duality $V^\ast\times V \to \mathbb{C}$ with the same ideas as above by looking at right ideals and the product of minimal right and left ideals, but that only gives upper and lower bounds on the topology of $V$ (this is the Mackey-Arens theorem).

This also answers the second question in the affirmative: If $A\cong End(W)$, then $V\cong W$ because both are isomorphic to the minimal left ideals contained in the minimal two-sided ideal.

thinkthis follows from an old result of Eidelheit, but since the conclusion of that result is usually stated as the weaker conclusion "$V\cong W$ as TVS" I'd need to look at the proof again in more detail $\endgroup$ – Yemon Choi Jan 21 '18 at 23:36Milutin's theorem$\endgroup$ – Yemon Choi Jan 23 '18 at 2:21