Long comment:
It should be pointed out that actually much less structure is needed to show that the separable $C^*$-algebras form a set than what Andreas Thom uses in his answer:
The crucial fact is that
there is a set of representatives of isometry classes of separable metric spaces.
This is essentially because separable metric spaces are of bounded size (see Komjath's comment),
namely of size $\leq 2^{\aleph_0}$.
Each separable metric space carries only a set of vector space structures over $\mathbb C$.
Each metric vector space over $\mathbb C$ only carries a set of binary and unary operations.
So, we obtain a set of representatives of the isomorphism classes of separable
$C^\ast$-algebras
without ever using the structure of $C^\ast$-algebras. Just the fact that they are separable metric spaces with a vector space structure over $\mathbb C$ and a fixed number of binary and unary operations.