We can consider the function
$$\displaystyle\sum_{k=1}^\infty{\frac{x^k}{1-y^k}} .$$
Is it possible to obtain a closed form expression for this? Or, if not, perhaps an integral is possible. Can we obtain any form of expression that states this without using summation? If so, I'd like to see how we can obtain this expression.
Alternatively, I'm also interested in the expression:
$$\displaystyle\sum_{k=1}^\infty{\frac{x^k}{1+y^k}} .$$
Your first sum can be expressed as a $q$-digamma function (the Lambert series being a special case of this); more precisely, we have (using sweetjazz's form):
$$\sum_{\ell=0}^\infty \frac{xy^\ell}{1-xy^\ell}=\frac{\psi _y^{(0)}\left(\frac{\log\,x}{\log\,y}\right)+\log(1-y)}{\log\,y}$$
The second series requires a bit more work; first we note that
$$\sum_{k=1}^\infty{\frac{x^k}{1+y^k}}=\sum_{\ell=0}^\infty (-1)^\ell \frac{xy^\ell}{1-xy^\ell}=\sum_{\ell=0}^\infty \frac{xy^{2\ell}}{1-xy^{2\ell}}-\sum_{\ell=0}^\infty \frac{xy^{2\ell+1}}{1-xy^{2\ell+1}}$$
The two series in the last expression now bear some resemblance to the series for the $q$-digamma function. In fact, we have the following:
$$\sum_{k=1}^\infty{\frac{x^k}{1+y^k}}=\frac{\psi_{y^2}^{(0)}\left(\frac{\log\,x}{2\log\,y}\right)-\psi_{y^2}^{(0)}\left(\frac12\left(1+\frac{\log\,x}{\log\,y}\right)\right)}{2\log\,y}$$
It looks to me that both series can alternatively be expressed as (products/ratios of) Jacobi theta functions (and their derivatives), but I haven't tried that route and my knowledge of modular identities is not up to snuff...