EDIT: Sorry - I did not think about threshold functions. I assumed we were working with a constant $p$, thought about the problem, and lost track of the actual question. I don't know how relevant it is, but I'll leave it.
Let $G$ be a bipartite graph on $2n$ vertices with parts $A$ and $B$ with $|A| = |B| = n$.
Let $E_A$ be the event that $A$ has an isolated vertex, $E_B$ be the event that $B$ has an isolated vertex. We're looking for $P(E_A \cup E_B) = P(E_A) + P(E_B) - P(E_A \cap E_B)$.
Each vertex in $A$ has $n$ possible edges, each with independent probability $p$, so the probability that it is isolated is $(1-p)^n$. Summing over the vertices of $A$, the probability that any vertex in $A$ is isolated is $n(1-p)^n$. The case for $B$ having an isolated is clearly symmetric. Hence, $P(E_A) = P(E_B) = n(1-p)^n$.
Now consider $P(E_A \cap E_B) = P(E_A)P(E_B|E_A)$. For any vertex in $B$, we have $n - 1$ possible edges (because we know one of the vertices in $A$ is isolated). The probability that a given vertex in $B$ is isolated is then $(1 - p)^{n - 1}$. Summing over the vertices of $B$, $P(E_B|E_A) = n(1-p)^{n-1}$. So $P(E_A \cap E_B) = (n(1-p)^n)(n(1-p)^{n-1}) = n^2(1-p)^{2n-1}$.
Putting it all together, $P(E_A \cup E_B) = 2n(1-p)^n + n^2(1-p)^{2n - 1}$.
As for a reference, I highly recommend Diestel's Graph Theory, which has an introductory treatment of random graphs (you may be above this level). There's a free preview here - http://diestel-graph-theory.com/. Bollobas' Modern Graph Theory has a similar but slightly deeper treatment. Bollobas also wrote a text called Random Graphs, which I know nothing about, but he is considered a leader in the field. I'm not sure that any of these address your specific problem.
The paper "Randomly Traceable Graphs" by Gary Chartrand and Hudson V. Kronk tackles this question, and confirms that the three types of graph ($C_n$, $K_n$, $K_{m,m}$) are the only ones:
https://epubs.siam.org/doi/abs/10.1137/0116056
Randomly Traceble graphs are graphs where every random self-avoiding path eventually visits all vertices (i.e. becomes a Hamiltonian path).
Randomly Hamiltonian graphs are Randomly Traceble graphs where every one of those random Hamiltonian paths can be closed to become a Hamiltonian cycle.
The paper first proves that all Randomly Traceble graphs with $n\ge3$ are randomly Hamiltonian. This is simple: Given any randomly traceable graph with a Hamiltonian path $v_1$, $v_2$, ..., $v_n$. The path $v_2$, ..., $v_n$, must be extendable to a Hamiltonian path since the graph is Randomly Traceable. Therefore $v_n$ and $v_1$ are connected by an edge, and any Hamiltonian path can be closed to become a Hamiltonian cycle. In the rest of the paper this trick of proving an edge must be present by constructing a hamiltonian path between its two vertices is repeatedly used.
The paper then views any Randomly Traceable graph as an outer (Hamiltonian) n-cycle possibly with diagonals.
If it has no diagonals, then it is just a cycle graph $C_n$.
If it has triangles (formed by two outer edges and one diagonal) then it turns out that all diagonals must be present and it is a complete graph $K_n$.
If instead it has 4-cycles (formed by three outer edges and one diagonal) then it turns out that it must be $K_{n/2,n/2}$ with $n$ even.
If it has larger cycles, then it will also have triangles, so no other types of randomly Hamiltonian graphs exist.
Best Answer
This problem is harder than finding an automorphism fixing two vertex associations.
Let $G(V, E)$ be a graph, and let $u_1, u_2, v_1, v_2 \in V$ be vertices such that we want to find if there is an automorphism bringing $u_1$ to $v_1$ and $u_2$ to $v_2$.
We replace every edge of $G$ with two arcs pointing in opposite directions. We replace each arc with an undirected gadget sufficient to ensure that arcs must be mapped to arcs. We thus get a new graph $\widetilde{G}$.
We define $e_1 = u_1v_1$ and $e_2 = u_2v_2$. $\widetilde{G}\cup e_1$ and $\widetilde{G}\cup e_2$ are isomorphic only if $G$ has an automorphism bringing $u_1$ to $v_1$ and $u_2$ to $v_2$.
It is easy to show that finding an automorphism fixing two vertex associations is harder than finding an automorphism.
Graph automorphism may be easier than graph isomorphism, but currently, we don't have asymptotically better algorithms for graph automorphism than for graph isomorphism. I was unable to show that this problem is complete for the Graph Automorphism class.
Note that it is necessary to build a $\widetilde{G}$ for the reduction since we can find some counterexamples such that $G\cup e_1$ and $G\cup e_2$ are isomorphic but have no isomorphism sending $e_1$ to $e_2$: