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.

Long comment:

It should be pointed out that actually much less structure than what Andreas Thom uses in his answer is needed to show that the isomorphism classes of separable $C^*$-algebras have a set of representatives:

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 at most $2^{\aleph_0}$. 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.

Note that I have never assumed that the metric, the vector space structure, and the additional operations interact in any way whatsoever.