Find a generating function for product of Stirling numbers. What is a close form of the series
$$
G_{i,j}(z)=\sum_{p=0}^\infty S(p,i) S(p,j) z^p? 
$$
Here $S(*,*)$ is the Stirling numbers of the second kind.
For $i=1$, since $S(p,1)=1$ the result is well-known 
$$
G_{1,j}(z)=\sum_{p=0}^\infty  S(p,j) z^p=\frac{z^j}{(1-z)(1-2z)\cdots(1-jz)}
$$
For small $i,j>1$ with computer I found  that 
$$
G_{2,2}(z)= -{\frac {z^2(2\,z+1)}{ \left( z-1 \right)  \left( 2\,z-1 \right)  \left( 4
\,z-1 \right) }},\\
G_{2,3}(z)=3\,{\frac {z^3(4\,{z}^{2}+2\,z-1)}{ \left( z-1 \right)  \left( 6\,z-1
 \right)  \left( 4\,z-1 \right)  \left( 3\,z-1 \right)  \left( 2\,z-1
 \right) }},\\
G_{2,4}(z)=-{\frac {z^4(24\,{z}^{2}+18\,z-7)}{ \left( z-1 \right)  \left( 6\,z-1
 \right)  \left( 4\,z-1 \right)  \left( 3\,z-1 \right)  \left( 2\,z-1
 \right)  \left( 8\,z-1 \right) }}
$$
What about the general case $G_{i,j}(z)?$ I have tried consider the $G_{i,j}(z)$ as Hadamar product $G_{1,i}(z) * G_{1,j}(z)$ but without any success.
 A: Starting from the EGF
$${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}$$
which is the labeled combinatorial class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}_{=k}(\textsc{SET}_{\ge 1}(\mathcal{Z}))$$
we obtain
$$\frac{1}{k!} n! [z^n] 
\sum_{p=0}^k {k\choose p} (-1)^{k-p} \exp(pz)
= \frac{1}{k!} 
\sum_{p=0}^k {k\choose p} (-1)^{k-p} p^n.$$
Documenting the choice of variables we also write
$${n\brace m} = \frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q^n.$$
We thus have for 
$$G_{k,m}(z) = \sum_{n\ge 0} {n\brace k} {n\brace m} z^n
\\ = \sum_{n\ge 0} z^n
\frac{1}{k!} 
\sum_{p=0}^k {k\choose p} (-1)^{k-p} p^n
\frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q^n
\\ = 
\sum_{n\ge 0} z^n \frac{1}{k!}
(-1)^k [[ n = 0 ]]
\frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q^n
\\ + \sum_{n\ge 0} z^n
\frac{1}{k!} 
\sum_{p=1}^k {k\choose p} (-1)^{k-p} p^n
\frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q^n.$$
The first term vanishes here and we continue with
$$\sum_{n\ge 0} z^n
\frac{1}{k!} 
\sum_{p=1}^k {k\choose p} (-1)^{k-p} p^n
\frac{1}{m!} (-1)^m [[ n = 0 ]]
\\ + \sum_{n\ge 0} z^n
\frac{1}{k!} 
\sum_{p=1}^k {k\choose p} (-1)^{k-p} p^n
\frac{1}{m!} 
\sum_{q=1}^m {m\choose q} (-1)^{m-q} q^n.$$
The first term again has a closed form and we obtain
$$\frac{(-1)^{k+m+1}}{k!\times m!} 
+ \sum_{n\ge 0} z^n
\frac{1}{k!} 
\sum_{p=1}^k {k\choose p} (-1)^{k-p} p^n
\frac{1}{m!} 
\sum_{q=1}^m {m\choose q} (-1)^{m-q} q^n.$$
We continue with the triple sum:
$$\frac{1}{k!} 
\sum_{p=1}^k {k\choose p} (-1)^{k-p}
\frac{1}{m!} 
\sum_{q=1}^m {m\choose q} (-1)^{m-q} 
\sum_{n\ge 0} z^n (pq)^n
\\ = \frac{1}{k!} 
\sum_{p=1}^k {k\choose p} (-1)^{k-p}
\frac{1}{m!} 
\sum_{q=1}^m {m\choose q} (-1)^{m-q} 
\frac{1}{1-pqz}.$$
Re-writing we find
$$\frac{1}{k!} \frac{1}{m!}
\sum_{l=1}^{km} \frac{1}{1-lz}
\sum_{p|l \wedge p\le k \wedge l/p \le m}
{k\choose p} (-1)^{k-p}
{m\choose l/p} (-1)^{m-l/p}.$$
Simplifying and collecting everything now yields
$$\frac{(-1)^{k+m+1}}{k!\times m!} 
+ \frac{(-1)^{k+m}}{k! \times m!} 
\sum_{l=1}^{km} \frac{1}{1-lz}
\sum_{p|l \wedge p\le k \wedge l/p \le m}
(-1)^{p+l/p}
{k\choose p} {m\choose l/p}.$$
The binomial coefficients control the  range, being zero when $p\gt k$
and / or $l/p \gt m$ and we may simplify even more to get
$$\bbox[5px,border:2px solid #00A000]
{-\frac{(-1)^{k+m}}{k!\times m!} 
+ \frac{(-1)^{k+m}}{k! \times m!} 
\sum_{l=1}^{km} \frac{1}{1-lz}
\sum_{p|l} (-1)^{p+l/p}
{k\choose p} {m\choose l/p}.}$$
We have  computed the  partial fraction  decomposition of  the desired
generating function  $G_{k,m}(z).$ Observe that this  will confirm the
three formulae  provided by  OP.
