In Flajolet and Sedgewick's Analytic Combinatorics pp. 261 (Chapter IV Note IV.28), the authors state that
\begin{align} [z^n]\ln \left(\frac{1}{1- \ln \frac{1}{1-z}}\right) \sim \frac{1}{n} (1-e^{-1})^{-n} \tag{1} \end{align}
where $S(z) = \ln \left(\frac{1}{1-\ln \frac{1}{1-z}}\right)$ is the EGF of the "super-necklaces", which is defined as
\begin{align} \mathcal{S} = CYC \circ CYC(\mathcal{Z}) \end{align}
I can't reach $(1)$ as $S(z)$ cannot be expressed as a rational function $\frac{f(z)}{g(z)}$. Also, the derivative of $S(z)$ still consists of the factor $\left(1-\ln \frac{1}{1-z}\right)$, which is $0$ at the pole $1-e^{-1}$. (A usual method to solve this kind of problems is to find the derivative $g'(z_0)$, where $z_0$ is the pole of $S(z)$. But it looks not applicable in this problem.)
Is there any method to solve the problem?
If you perform the expansion near $z\to \rho$ you end up with (if I'm not mistaken) $S(z) = \ln(\frac 1e) + \ln(\frac 1 {\rho-z}) + O(\rho-z)$.
I obtained this by simply setting $h = \rho-z$ then using Taylor expansion of $\ln$ keeping in mind that $z\to 0$. Namely
$$S(z) = \ln(\dfrac 1 {1- \ln(\frac {1}{e^{-1} +h})}) = \ln(\dfrac 1 {1-\ln e + \ln(1+eh)}) = \ln(\dfrac 1 {\ln(1+eh)}) = \ln(\dfrac 1 {eh + O(h^2)}) = \ln(\frac 1e) + \ln(\frac 1 h)+\ln(\frac 1 {1+O(h)}) = QED$$ The results of chapter VI (transfer theorems) allow you to work such problems out when you don't have an exactly meromorphic function anymore but only an asymptotic estimate. Basically your coefficients will be asymptotic to the coefficients of the leading term.
There are some technical conditions ($\Delta$-analyticity) that you have to check, which is not a problem in your case, where your function should be defined on a slit disk).