Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?
Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it.
No. Take $M$ to be any II$_1$-factor. Let $\pi:M\to B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $\pi$ is injective. And $\pi(M)$ is dense in $B(H)$, but it cannot be everything.
So $\pi:M\to\pi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.
As mentioned in the comments, if $M\subset B(H)$ and $N\subset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $\pi:M\to N$ is sot-continuous on bounded sets by passing through normality.