I am writing here since I cannot comment yet. Please note that, if I am not mistaken, for the case of $q \in (0,1)$ the sum diverges, since
$$ \sum_{x=1}^{\infty} {x^a\over 1-q^{x}} = \sum_{x=1}^{\infty} \sum_{l=0}^{\infty}{x^a q^{l x}} > \sum_{x=1}^{\infty} x^a = \infty $$
About your question concerning JJacquelin's manipulations, the reason he integrates is because he wants to get rid of the term $1/m$ in the denominator that prevents him from using his previous result. Doing this integration yields
$$ \int \frac{1}{x} \sum_{m=1}^\infty \frac{(x y)^m}{1-y^m} dx = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int x^{m-1}dx= \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m} = \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m} $$
If it had $m!$ in the denominator, as you ask, the same manipulation would have led to multiple integrations
$$ \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m!} = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m!} = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \sum_{m=1}^\infty \frac{y^m}{1-y^m}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \frac{1}{x_1^m}\sum_{m=1}^\infty \frac{(x_1 y)^m}{1-y^m}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \frac{1}{x_1^m}\left [ \psi_y \left( 1 + \frac{\ln(x_1)}{\ln(y)} + \ln(1-y) \right) \right]dx_1 dx_2 \dots dx_m $$