The vast majority of disconnected graphs have a single isolated vertex.
Let $A$ be a nonempty proper subset of $\{1,...,n\}$ of size $a$. Let $s(a)$ be the number of graphs with
$e=\lfloor \frac12 {n \choose 2}\rfloor$ edges which have no edges from $A$ to $A^c$.
We want to count the union of all of these. Inclusion-exclusion works, with the dominant terms coming from when $a=1$.
An upper bound is the sum of $s(a)$ over all $A$ of size at most $n/2$, which is at most
$n ~s(1)$ + ${n\choose 2}s(1)$ + $2^ns(3)$.
To get a lower bound, subtract the number of graphs with no edges connecting $A$ to $A^c$ or edges connecting $B$ to $B^c$ for all disjoint $\{A,B\}$. Denote this by $s(\#A,\#B)$. So, subtract
${n\choose2}s(1,1) + 3^ns(1,2)$ from $n~s(1)$.
The rest should be routine estimates on $s(1)$, $s(2)$, $s(3)$, $s(1,1)$, and $s(1,2)$.
$s(a,b) \le s(a+b)$.
$s(a) = ({n\choose 2} -a(n-a))$ choose $e$.
Let the total number of graphs with $e$ edges be $\#G = s(0)$.
$$s(a)/\#G = \prod_{i=0}^{a(n-a)-1} \frac{\lceil{n\choose2}/2\rceil-i}{{n\choose2}-i}.$$
$s(2)/s(1) \le 2^{-n+3}$.
$s(3)/s(1) \le 2^{-2n+8}$.
The dominant term in both the upper bound and the lower bound is $n~s(1)$.
If I calculated correctly, that's asymptotic to $\frac 2 e n 2^{-n} ~\#G$.
I'm not sure if this is the right interpretation or not...it may really just be another way of encoding the generating function argument. Let $H$ be a random bipartite graph where every edge appears independently with probability $1/2$. Then the left hand side is
$$2^{n^2} E \left(\prod_v f(v) \right),$$
where $f(v)$ is equal to $\sum_u x(u,v)$ and $x(u,v)$ is $1$ if an edge is not present, $-1$ if an edge is present. Expanding out the product and using linearity of expectation, we can write this as
$$2^{n^2} \sum_{\sigma} E \left(\prod_{v} x(v,\sigma(v))\right)$$
Where $\sigma$ consists of all mappings taking each vertex to a vertex on the opposite side.
Any $\sigma$ for which some edge $(v,\sigma(v))$ appears only once has $0$ expectation due to independence. The $\sigma$ for which every edge appears for both of its endpoints correspond to matchings between the left and right side, of which there are $n!$. (This last observation corresponds to the fact that the expected square of the permanent of an $n \times n$ random Bernoulli matrix is $n!$...I think it goes at least back to Turan).
Best Answer
It seems that what Andrew wants to count are what are called in enumerative contexts "bicolored graphs". A bicolored graph is a graph in which the vertices have been colored black and white so that every edge joins two vertices of different colors. A bipartite (or bicolorable) graph is a graph that has a bicoloring. A bicolorable graph with $k$ connected components has $2^k$ bicolorings. (In nonenumerative contexts the distinction between bipartite and bicolored is usually unimportant.) In addition, in counting bicolored graphs one might or might not consider switching the two colors to give an equivalent graph. All of the versions of the enumeration problem have been solved. Counting unlabeled bicolored graphs (with no color-switching equivalence) is a straightforward application of Burnside's lemma; counting unlabeled bipartite graphs is tricky.
It's not too hard to find appropriate references by searching MathSciNet. (Hint: "color" is sometimes spelled "colour".)
Incidentally, the number of labeled bicolored graphs on $n$ vertices is $$\sum_{m=0}^n 2^{m(n-m)}\binom{n}{m}.$$