Denote $d_A(v)=|N_G(v)\cap A|$, $d_B(v)=|N_G(v)\cap B|$. Set $S_A=\{(a,b)\in A\times B\colon N_G(a)\cap N_G(b)\cap A\neq\varnothing\}$ and $S_B=\{(a,b)\in A\times B\colon N_G(a)\cap N_G(b)\cap B\neq\varnothing\}$. Clearly, $S_A,S_B\subseteq S$. Now we claim that $|S_A|\geq 2|E(G[A])|$ and, similarly, $|S_B|\geq 2|E(G[B])|$; this yields the required inequality.

To prove the first inequality, consider any edge $e=(a_1,a_2)$ with $a_1,a_2\in A$; denote $d_i=d_B(a_i)$ and notice that $d_i\geq d_A(a_i)\geq 1$. For each neighbor $b_1\in B$ of $a_1$ (there are $d_1$ such), we say that $e$ contribures $1/d_2$ to the pair $(a_2,b_1)\in S_A$. Similarly, for every neighbor $b_2$ of $a_2$ (there are $d_2$ such), the edge $e$ contributes $1/d_1$ to the pair $(a_1,b_2)\in S_A$. Thus, the total contrivution of $e$ is $d_1/d_2+d_2/d_1\geq 2$, and the total contribution of all edges is $\geq 2|E(G[A])|$. On the other hand, the total contribution to any pair $(a,b)\in S_A$ is at most $d_A(a)/d_B(a)\leq 1$. Hence, the total contribution of all edges is at most $|S_A|$. Therefore, $|S_A|\geq 2|E(G[A])|$, as required.