I guess no?
Let $A$ be commutative and let $B$ be $C(X)\subseteq B(l^2(X))$, where $X$ is compact Hausdorff. Suppose $B$ is weakly closed, take an arbitary $x\in X$, there is a net of operators in $B$ which weakly converges to $1_{\{x\}}$:
$$\mathscr Y=\{Y\subseteq_{finite} X\backslash \{x\}\}\\f_Y(y)=\left\{\begin{array}{}1&,y=x\\0&,y\in Y\\ others&,y\notin Y\cup \{x\}\end{array}\right.$$
Therefore $1_{\{x\}}\in B$. However, $A$ may have no minimal projections.
Is this counter example correct?
But isn't it kind of wired that being isomorphic to a Von Neumann algebra doesn't make itself a Von Neumann algebra?
I think your counter-example is not correct since your initial assumptions ended up implying that $X$ is discrete $\big (1_{\{x\}}\in B\big )$, hence $X$ is finite, so $C(X)$ is a von Neumann algebra, after all.
According to the original definition of von Neumann algebras it doesn't make sense to say that "$M$ is a von Neumann", rather one would say that "$M$ is von Neumann algebra of operators on $H$", where $H$ is a Hilbert space. In 1956 Sakai found an intrinsic characterization of von Neumann algebras according to which a C*-algebra $A$ admits a faithful representation on some Hilbert space $H$ whose range is a weakly closed algebra of operators on $H$ iff $A$ is a dual Banach space.
Sakai's result may then be interpreted as a definition of von Neumann algebras which is representation independent.
I believe your question can be interpreted as asking whether or not an abstract von Neumann algebra admits a faithful representation whose range is not a von Neumann algebra.
It is not hard to produce examples of such representations for any given infinite-dimensional von Neumann algebra $M$. All you have to do is choose a non-normal state and consider the corresponding GNS representation, adding to it a copy of any old faithful representation to make the whole thing faithful. If $\xi $ is the cyclic vector, then the map $$ T\in M\mapsto \langle T\xi , \xi \rangle \in \mathbb C $$ will fail to be normal, attesting to the fact that $M$ is not a von Neumann in that representation.