Coefficients of a formal power series satisfying $\exp(f(z)) = 1 + f(q\,z)/q$ Let $(q;\,q)_n$ denote the $q$-Pochhammer symbol:
$$(q;\,q)_n = \prod_{k=1}^n (1 - q^k), \quad(q;\,q)_0 = 1.\tag1$$
Consider a formal power series in $z$:
$$f(z) = \sum_{n=1}^\infty \frac{(-1)^{n+1}P_n(q)}{n!\,(q;\,q)_{n-1}}z^n,\tag2$$
where $P_n(q)$ are some (yet unknown) polynomials in $q$: 
$$P_n(q) = \sum_{k=0}^{m} c_{n,k} \, q^k,\tag3$$
where $m=\binom{n-1}2 = \frac{(n-1)(n-2)}2$ and $c_{n,k}$ are some integer coefficients.
Suppose the formal power series $f(z)$ satisfies the functional equation
$$\exp(f(z)) = 1 + f(q\,z)/q.\tag4$$
Expanding the left-hand side of $(4)$ in powers of $z$ using the exponential partial Bell polynomials, and comparing coefficients at corresponding powers of $z$ at both sides, we can obtain a system of equations, by solving which we can find the coefficients of the polynomials $P_n(q)$:
$$
\begin{align}
P_1(q) &= 1\\
P_2(q) &= 1\\
P_3(q) &= 2 + q\\
P_4(q) &= 6 + 6 q + 5 q^2 + q^3\\
P_5(q) &= 24 + 36 q + 46 q^2 + 40 q^3 + 24 q^4 + 9 q^5 + q^6\\
\dots
\end{align}\tag5
$$
This is a quite slow process, even when done on a computer. I computed the polynomials up to $n=27$ (they can be found here) using a Mathematica program that can be found here.
There are some patterns in the coefficients I computed (so far they are just conjectures):
$$
\begin{align}
c_{n,0} &= (n-1)!&\vphantom{\Huge|}\\
c_{n,1} &= \frac{(n-2)(n-1)!}2, &n\ge2\\
c_{n,2} &= \frac{(3n+8)(n-3)(n-1)!}{24}, &n\ge3\\
c_{n,3} &= \frac{(n^2 + 5 n - 34)\,n!}{48}, & n\ge4
\end{align}
\tag6
$$
and
$$
\begin{align}
c_{n,m} &= 1&\vphantom{\Huge|}\\
c_{n,m-1} &= \frac{(n+1)(n-2)}2, &n\ge2\\
c_{n,m-2} &= \frac{(3 n^3 - 5 n^2 + 6 n + 8)(n-3)}{24}, &n\ge3\\
c_{n,m-3} &= \frac{(n^4 - 10 n^3 + 43 n^2 - 74 n + 16) (n - 1) \, n}{48}, &n\ge4
\end{align}
\tag7
$$
where $m=\binom{n-1}2$. Other coefficients seem to follow more complicated patterns. We can also observe that
$$
\begin{align}
P_n(1) &= \frac{(n-1)!\,n!}{2^{n-1}}\\
P_{2n}(-1) &= \frac{(2n-1)!\,n!}{3^{n-1}}\\
P_{2n-1}(-1) &= \frac{(2n-1)!!\,(2n-2)!}{6^{n-1}},
\end{align}\tag8
$$
where $n!!$ denotes the double factorial.
I am trying to find a more direct formula for the polynomials $P_n(q)$ or their coefficients $c_{n,k}$ (possibly, containing finite products and sums, but not requiring to solve equations).

Using the first recurrence relation here, we can find a recurrence for the polynomials $P_n(q)$:
$$P_1(q) = 1, \quad P_n(q) = \sum_{k=1}^{n-1} {{n-1} \choose {k-1}} {{n-2} \brack {k-1}}_q P_k(q) \, P_{n-k}(q) \, q^{n-k-1},$$
where $n \choose k$ is the binomial coefficient, and ${n \brack k}_q$ is the $q$-binomial coefficient (also known as the Gaussian binomial coefficient). A Mathematica program that computes them using this recurrence can be found here.
If we introduce a notation for the coefficients of the formal power series $f(z)$, that are rational functions of $q$:
$$f(z) = \sum_{n=1}^\infty Q_n(q)\,z^n, \quad Q_n(q) = \frac{(-1)^{n+1}P_n(q)}{n!\,(q;\,q)_{n-1}},$$
then we can have a simpler recurrence for them:
$$Q_1(q) = 1, \quad Q_n(q) = \frac1{n \, (1-q^{1-n})}\sum_{k=1}^{n-1} k \, q^{-k} \, Q_k(q) \, Q_{n-k}(q).$$
It would be nice to find a more direct, non-recurrent formula for them.
