Question

Your argument is correct. An alternate and more "intrinsic" argument is to look at the predual, which is a separable Banach space. There are only continuum many separable Banach spaces, since they are determined by the metric on a countable dense subset that is a $\mathbb Q[i]$-vector space. Since a von Neumann algebra is determined by its predual, this implies Thus there are only continuum many separable von Neumann algebras, when considered just with their structure of dual Banach spaces. As Nik points out in the comments, the dual Banach space structure may not determine the algebra structure. However, given any such dual Banach space, you can fix a countable weak*-dense subset, and then any algebra structure will be determined by what it does on that countable set, so you again have only continuum many possibilities.