Here is a proof of the formula 
$$\sum_{j>0} \frac{1}{j} w(q)^j P_qw(q)^{-j}=2\log \frac{w(q)}{q}-\log w'(q).$$
(The  notation is the same as in the Timothy's answer except I prefer $P_qf(q)$ to $[q^{>0}]f(q)$.) Consider $q$ fixed. We may assume without loss of generality that $w(z)$ is analytic in some circle $|z|\le R$ 
where $R>|q|$. We assume additionally that it is univalent in this circle and that $|w(z)|>|w(q)|$ when $|z|=R$. 

We have a formula 
$$P_q f(q)=\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)f(z)\,dz$$
which is valid when $f$ is analytic in the punctured circle $0<|z|\le R$. 
A straightforward calculation then gives 
$$\sum_{j>0} \frac{1}{j} w(q)^j P_qw(q)^{-j}=\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)}{w(z)-w(q)}\,dz.$$ 
(Importantly, the series converges uniformly so there are no analytic issues.) This  integral is problematic because of branching points of the  logarithm but there is a trick to circumvent this obstacle, 
$$\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)}{w(z)-w(q)}\,dz=\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)(z-q)}{z(w(z)-w(q))}\,dz+\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{z}{z-q}\,dz.$$
Under close examination, 
$$\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{z}{z-q}\,dz=0.$$
(I found it out  using the dilogarithm but there may be better proofs.) The logarithm  in the first integral is now analytic in the whole circle so we can easily compute it with  residues, 
$$\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)(z-q)}{z(w(z)-w(q))}\,dz=2\log\frac{w(q)}{q}-\log w'(q).$$
This is it.