Since you've asked for a hint rather than a complete solution I'm giving a hint. I think this will lead to complete solution, but I haven't worked out the details, which I'm leaving to you. Of course it's possible that this doesn't work.

This is based on Timothy Budd's formulation.

Let us look at a slightly different but closely related problem. Given a power series $G(t)$ with constant term 1, find
$$\sum_{j>0}\frac{1}{j} x^j [q^{>0}] \left(\frac{G(q)}{q}\right)^{j}\tag{1}$$

If we can solve this problem then we can just replace $G(q)$ with $q/w(q)$. and then replace $x$ with $w(q)$.

The sum in $(1)$ is equal to
$$
[q^{>0}]\log\left(\frac{1}{1-xG(q)/q}\right)
$$
(Here we are working in $\mathbb{C}((q))[[x]]$, or perhaps $\mathbb{C}[[q,x/q]]$—we allow negative powers of $q$ but not of $x$.)

Now let $g(x)$ be the formal power series solution of $g(x) = x G(g(x))$. 
Then have the formulas
$$
[q^{<0}]\log\left(\frac{1}{1-xG(q)/q}\right)=\log\left(\frac{1}{1-g(x)/q}\right) \tag{2}
$$
and
$$
[q^{=0}]\log\left(\frac{1}{1-xG(q)/q}\right)=\log\left(\frac{g(x)}{x}\right) \tag{3}$$

Equation (2) is easily seen to be equivalent to ordinary Lagrange inversion. Equation (3) is not so well known, but it's easily derived from well-known forms of Lagrange inversion. It can be found, for example, in my paper on Lagrange inversion[ \[I. M. Gessel, Lagrange inversion, J. Combin. Theory Ser. A 144 (2016), 212–249\]][1], equation (2.2.9).

Therefore 
$$\begin{aligned}
\ [q^{>0}]\log\left(\frac{1}{1-xG(q)/q}\right) &= \log\left(\frac{1}{1-xG(q)/q}\right) 
  -\log\left(\frac{1}{1-g(x)/q}\right)\\ &\qquad\qquad-\log\left(\frac{g(x)}{x}\right).
  \end{aligned}
$$


  [1]: https://doi.org/10.1016/j.jcta.2016.06.018