Karp hardness of searching for a matching split
It is NP-complete. Clearly it belongs to NP. To prove its NP-completeness, we'll reduce not-all-equal 3-satisfiability (NAE3SAT) to this problem.

Given an instance of NAE3SAT with $n$ variables and $m$ clauses, construct a graph as follows:


For each literal $l$, construct a complete graph with $m+1$ vertices $v_0(l),\ldots,v_m(l)$.
For each variable whose positive and negative literals are respectively $l_i$ and $l_j$, add an edge between $v_0(l_i)$ and $v_0(l_j)$.
For each clause $c_i=l_i^1\vee l_i^2\vee l_i^3$, add the following structure:


In addition, for each literal $l$ other than $l_i^1,l_i^2,l_i^3$, add a new vertex $v_i'(l)$ and an edge between $v_i(l)$ and $v_i'(l)$.


Now we assert that the instance of NAE3SAT is satisfiable if and only if the graph has a matching split.

If the instance of NAE3SAT is satisfiable, consider a valid truth assignment. If $l$ is a truth literal, we put $v_i(l)$'s into $A$ and $v_i'(l)$'s (if exists) into $B$ for all $i=0,\ldots,m$, otherwise we put $v_i(l)$'s into $B$ and $v_i'(l)$'s into $A$. In addition, for the structure corresponding to $c_i$ (we assume $l_i^1$ and $l_i^2$ are truth literals while $l_i^3$ is a false literal without loss of generality), we put $v_i^{12}, v_i^{123}$ into $A$ and put $v_i^{23}, v_i^{31}, v_i^{122331}$ into $B$. We can see this partition is a matching split.

If the graph has a matching split, we can observe firstly that $v_0(l),\ldots,v_m(l)$ should be put into the same part since they make up a complete graph (assume $m\ge 2$ without loss of generality). In addition, if $l_i$ and $l_j$ correspond respectively to the positive and negative literals of the same variable, $v_k(l_i)$ and $v_k(l_j)$ should be put into different parts otherwise $v_0(l_i)$ or $v_0(l_j)$ has no neighbor in the other part. If the positive literal corresponding to a variable is put into $A$, we assign the variable to True, otherwise we assign it to False. Note for each clause $c_i$, $v_i(l_i^1), v_i(l_i^2)$ and $v_i(l_i^3)$ cannot be put into the same part otherwise $v_i^{123}$ has either no neighbor or $3$ neighbors in the other part. Hence this assignment is a valid assignment for the instance of NAE3SAT.



Edit: One can simply add edges among $v_0(l)$'s to make the graph connected. For example, if the clauses are $l_1\vee l_2\vee l_3$ and $l_4\vee l_5\vee l_6$, one can simply add an edge between $v_0(l_3)$ and $v_0(l_4)$. The idea is that for a valid assignment, if we flip all the values of variables involved in one connected component, the assignment is still valid. In the example, this means we can always assume $v_0(l_3)$ and $v_0(l_4)$ are put into the same part.