Now that Yemon has given a counterexample, I thought it worth giving a stronger hypthoesis under which we have a positive answer.

I looked at something similar here: http://arxiv.org/abs/math/0604372 see Section 4.

**Theorem:** Let $M$ be a commutative von Neumann algebra, and let $N$ be a dual Banach algebra. Any bounded algebra isomorphism between $M$ and $N$ is automatically weak$^*$-continuous.

With Le Pham and White we remove the commutative condition in http://arxiv.org/abs/0804.3764

With ten years' of hindsight, I find that I can now improve this result and give an easy proof... A dual Banach algebra is a Banach algebra which is a dual space, in such a way that the multiplication is separately weak$^*$-continuous. Thus the above result says that a von Neumann algebra has a unique weak$^*$-topology making the multiplication map weak$^*$-continuous.

If $E$ is a Banach space, then whenever $E$ is isomorphic to $F^*$ for some Banach space $F$, say under $\theta: E \rightarrow F^*$ we can look at the image of $F$ under the adjoint $\theta^*$. This gives a closed subspace of $E^*$, say $X$, such that $E$ is canonically isomorphic to $X^* = E^{**} / X^\perp$, where $X^\perp$ is the annihilator of $X$ in $E^{**}$. This means that:

- For each $0\not=x\in E$ there is $\mu\in X$ with $\mu(x)\not=0$;
- For each $\Phi\in E^{**}$ there is $x\in E$ with $\mu(x) = \Phi(\mu)$ for all $\mu\in X$.

I call $X$ a "concrete predual" of $E$. The $x\in E$ given by (2) is unique, because of condition (1). Let $\alpha : E^{**} \rightarrow E$ be the map so given. It is a projection of $E^{**}$ onto $E$, once we canonically identify $E$ with its image in $E^{**}$.

**Theorem:** Let $A$ be a C$^*$-algebra which is isomorphic to a dual space $F^*$, with the additional assumption that $F$ is a Banach $A$-bimodule, and the isomorphism $A \cong F^*$ is a bimodule map. Then $A$ is a von Neumann algebra (and the weak$^*$-topology is the canonical one).

To prove this, argue as above to reduce to a concrete predual $X\subseteq A^*$. The bimodule condition becomes that $X$ is $A$-invariant for the usual action of $A$ on $A^*$. One can check that $X$ being $A$-invariant is actually equivalent to the map $\alpha:A^{**}\rightarrow A$ being an algebra homomorphism. Thus $\ker\alpha$ is a closed 2-sided ideal in $A^{**}$ and so is $*$-closed. This in turn is equivalent to $X$ being $*$-closed, which in turn implies that $\alpha$ is actually a $*$-homomorphism. Then $\alpha$ is a contractive projection which implies that $A$ is *isometrically* isomorphic $X^*$, hence a von Neumann algebra.

(This proof is obviously influenced by Tomiyama's proof that W$^*$-algebras and von Neumann algebras are the same thing.)