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.
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.