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. 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], equation (2.2.9).
Therefore $$\begin{aligned} [q^{>0}]\log\left(\frac{1}{1-xG(q)/q}\right) &= \log\left(\frac{1}{1-xG(q)/q}\right) -\log\left(\frac{1}{1-g(x)/q}\right)\\ &\qquad\qquad-\log\left(\frac{g(x)}{x}\right). \end{aligned} $$