A combinatorial identity on even spanning subgraphs in the Erdös-Renyi random graph with relations to the Ising model Let $x \in \lbrack 0,1 \rbrack$. Then for any finite graph $G$ consider the Erdös-Renyi random graph where we independently keep each of the edges with probability $x$. Denote the corresponding probability measure $P_x$. An even spanning subgraph of $G$ is a subgraph where all the vertices are kept and some of the edges such that each vertex has even degree. Denote the set of even spanning subgraphs of G by $\text{ess}(G)$.  For an even spanning subgraph $g$ let $O(g)$ be the event that all the edges are kept. 
By some knowledge about geometric representations of the Ising model I happen to know that for every pair of vertices $v_1,v_2 \in G$ we have 
\begin{align*}
\sum_{g_1, g_2 \in \text{ess}(G)} \frac{ P_{x^2} \left( \{v_1 \leftrightarrow v_2 \}, O(g_1), O(g_2) \right)}{P_x(O(g_1))}   = \sum_{g_1, g_2 \in \text{ess}(G)} P_x( \{v_1 \leftrightarrow v_2 \}, O(g_1)) P_x( \{v_1 \leftrightarrow v_2 \}, O(g_2)) 
\end{align*}
Here $\{v_1 \leftrightarrow v_2 \}$ is the event that there is a path of kept edges in $G$ connecting $v_1$ and $v_2$.
However, this is a problem purely on Erdös-Renyi random graphs, but I can't find a combinatorial proof of this which could maybe give some insights on how the relation generalizes beyond the class of events $\{ v_1 \leftrightarrow v_2 \}$.
 A: (warning: below $\{v_1 \leftrightarrow v_2 \}$ has different meaning than in OP)
Let me prove a slightly different identity
\begin{align*}
x\sum_{g_1, g_2 \in \text{ess}(G)} \frac{ P_{x^2} \left( \{v_1 \leftrightarrow v_2 \}, O(g_1), O(g_2) \right)}{P_x(\{v_1 \leftrightarrow v_2 \},O(g_1))}   = \\\sum_{g_1, g_2 \in \text{ess}(G)} P_x( \{v_1 \leftrightarrow v_2 \}, O(g_1)) P_x( \{v_1 \leftrightarrow v_2 \}, O(g_2)).
\end{align*}
If $E_i$ denotes edge set of $g_i$ and $e=\{ v_1 \leftrightarrow v_2 \}$, then LHS may be written as $$\sum_{g_1,g_2} x^{2|E_1\cup E_2\cup e|-|E_1\cup e|+1}=\sum_{g_1,g_2} x^{|E_2\cup e|+|(E_1\Delta E_2)\cup e|},$$
where we apply the general identity $2|A\cup B|-|A|=|B|+|A\Delta B|$ for $A=E_1\cup e$, $B=E_2\cup e$, and $|(E_1\Delta E_2)\cup e|=1+|A\Delta B|$. It remains to change the pair $(g_1,g_2)$ in RHS of my identity onto pairs $(g_1\Delta g_2,g_2)$.
I am still not sure what exactly do you mean (there are several typos at least), but possibly something provable along these lines.
