What's the inverse of the Weierstrass-Mittag-Leffler-Transform $\exp\left[g(z) + \int_\mathbb C f(y)\ln(z-y)\,dy\right]$?

Question

As mentioned in another post, as a consequence of Mittag-Leffler's theorem combined with the Weierstrass factorization theorem, after reducing to the common denominator, any meromorphic function can be expressed as

$$f(z) = e^{g(z)}\prod_k(z-z_k)^{n_k} \tag{1}\label{1}$$

where $n_k > 0$ ($n_k<0$) denotes the multiplicity of a zero (pole) at $z_k$ and $g(z)$ is an entire function.

This can be reformulated and generalized into

$$\begin{align} f(z) &= \exp\left[g(z) + \sum_k n_k\ln(z-z_k)\right] \\ &\to \exp\Big[g(z) + \underbrace{\int_\mathbb C n(y)\ln(z-y)\,dy}_{=(n\ast_\mathbb C\ln)(z)}\Big] \tag{2}\label{2} \end{align}$$

where $n(z)$ is a strictly integer-valued distribution for meromorphic functions, but could in general also be something else, e.g. a meromorphic function itself. Due to its origin I'd like to refer to $\eqref{2}$ as the Weierstrass-Mittag-Leffler-Transformation (or WeMiLe-Transformation, if you allow for that acronym). So now my question is

How can the transformation $\eqref{2}$ be inverted?

For a meromorphic $f(z)$ with finitely many zeros of finite order that inversion is obviously $n(y) = \sum_k n_k \delta(y-z_k)$ and $g(z)=\ln\big[f(\zeta)/\prod_k(\zeta-z_k)\big]$ for any $\zeta\in\mathbb C\backslash\{z_k\}$, i.e. one has to determine all zeros and poles, but what about non-meromorphic $f(z)$, e.g. what about $f(z)=\delta(z)$?

Answer 1

first attempt at a partial answer, to be improved...

I've noticed the exponent in $\eqref{2}$ is proportional to the antiderivative of the Hilbert-Transform of $n(z)$ if all zeroes and poles lie on the real axis, so formally this means

$$n(z) = -\frac1\pi\mathcal H \frac{d}{dz}\ln\frac{f(z)}{c} = -\frac1\pi\mathcal H \frac{f'(z)}{f(z)} \tag{n}\label{n}$$

where $\mathcal H$ denotes the Hilbert transform $\mathcal H f(z) = \Big(\frac1{\pi z}\ast f\Big)(z)$. But how valid is this really?

Ok, let's have some examples:

$f(z) = c$

Clearly $f'(z)=0=n(z)$

$f(z) = z - z_0$

So $f'(z) = 1$ and therefore \begin{align} n(z) &= -\frac1\pi\mathcal H \frac1{z-z_0} \\ &= \delta(z-z_0) \end{align}

So far, so good, though I didn't require $z_0\in\mathbb R$ here at first glance...

to do: do some more examples