Is the Banach Algebra generated by a separable subset separable?

Yes, it's true, and your proof idea works. The Banach algebra generated by $S$ is the closure of the (noncommuting) polynomials in $S$ as you say. So it suffices to show that any polynomial in $S$ (with complex coefficients) can be approximated by a sequence of polynomials in $D$ (with coefficients in $\mathbb{Q}(i)$), since then the closure of the latter contains the former and hence the closure of the former.

By linearity (and the continuity of addition and scalar multiplication), together with the density of $\mathbb{Q}(i)$ in $\mathbb{C}$, it suffices to show that a monomial $\prod_{i=1}^n s_i, s_i \in S$ can be approximated by a sequence of monomials in $D$. This follows from the continuity of multiplication: take for each $s_i$ a sequence $d_{i, j} \in D$ converging to $s_i$ and consider the sequence of monomials $j \mapsto \prod_{i=1}^n d_{i, j}$.