The Abel's identity which is given with Abel polynomials:
$$\sum_{k=0}^n\binom{n}{k}a(a+kz)^{k-1}b(b+(n-k)z)^{n-k-1}=(a+b)(a+b+nz)^{n-1}$$
A possible proof is to simply calculate coefficient of $x^n$ of $e^{(a+b)R(x)}$, where the $R(x)$ is a formal series satisfying $R(x)=xe^{zR(x)}$.
However, a more conbinatorial/intuitive proof is requested.
What I have tried but did not work (using generating functions):
$R(x)$ is the EGF of rooted labeled trees, $z$ is an intermediate tracking the number of children.
$e^{cR(x)}$ is the EGF of a forest of rooted labeled trees, $c$ is an intermediate tracking the number of rooted trees in a forest.
The proof from here is by using the fact that any forest of size $n$ can be constructed from two smaller forest of size $k$ and $n-k$.
The problem here is that the parameters cannot match. The RHS uses $a+b$ to track something the LHS uses $a$ and $b$ separately.
Is this method viable for a proof, or there's a better way?
Also, I cannot find a way to connect this identity to other ones, like:
$$\sum^n_{k=0}\binom{n}{k}x(x-kz)^{k-1}(y+kz)^{n-k}=(x+y)^n$$
Is there also a combinatorial proof for this?