This is not an answer but a conjecture. It based on a first examples examples calculated by Mathematica. $$S(a)=\frac1{a+1}-\sum_{i=1}^a\frac{\psi(-\alpha_i)}{f_a'(\alpha_i)},$$ where $f_0(x)=1$, $f_1(x)=x+3$, $f_2(x)=x^2+5x+7$,... $$f_{n+1}(x)=(x+2)f_n(x)+1\qquad(n\ge 0),$$ (see A193844) and $\alpha_i$ are the roots of $f_a$.

This is not an answer but a conjecture. It based on a first examples examples calculated by Mathematica. $$S(a)=\frac1{a+1}-\sum_{i=1}^a\frac{\psi(-\alpha_i)}{f_a'(\alpha_i)},$$ where $f_0(x)=1$, $f_1(x)=x+3$, $f_2(x)=x^2+5x+7$,... $$f_{n+1}(x)=(x+2)f_n(x)+1\qquad(n\ge 0),$$ (see A193844) and $\alpha_i$ are the roots of $f_a$.

UPD: From Fedor's formula $f_{n}(x)=\frac{(x+2)^{n+1}-1}{x+1}$ follows that for $\alpha_k=w_k-2$ we have $$f_n'(\alpha_k)=\frac{n+1}{w_k(w_k-1)}.$$