Let $A$ and $B$ be any two infinite-dimensional von Neumann algebras, they are operator algebras with operator composition as the multiplication and as infinite dimensional vector spaces they're isomorphic $F: A \rightarrow B$ right? I wanted to know if, or when are they isomorphic as algebras?? (Not necessarily *-isomorphic). That is, if they also satisfy
$F(ab) = F(a)F(b)$ ??
No. This has more to do with simple/non simple algebras in general, than just with von Neumann algebras.
Denote $B(H)$ the algebra of bounded linear operator on the infinite-dimensional separable Hilbert space $H$. Then take $$ A=B(H)\quad\mbox{and}\quad B=B(H)\oplus B(H) $$ These are easily seen to be isomorphic as complex vector spaces. But not as complex algebras.
To see that, observe that $B$ has two non-trivial central idempotents $(\rm{Id},0)$ and $(0,\rm{Id})$, While the only central idempotents of $A$ are trivial, i.e. $0$ and $\rm{Id}$.
If you prefer, the center of $A$ is $\mathbb{C}I$ while the center of $B$ is $\mathbb{C}I\oplus\mathbb{C}I$. If there was an algebra isomorphism between $A$ and $B$, it would induce an isomorphism between their centers. And they are not isomorphic for obvious dimension reasons.