There are no good exact answers.
For values of $N$ close to $M$, we can write down something intelligent. For example, when $N=M-1$, the total number of connected graphs with $M-1$ edges and $M$ vertices is $M^{M-2}$ (the number of labeled trees), so the probability is
$$
\frac{M^{M-2}}{M^{2M}} = \frac1{M^{M+2}}.
$$
(Assuming that you allow loops - edges with $i=j$ - which you seem to. Also, it's very confusing that you are using $M$ to denote vertices and $N$ to denote edges rather than the other way around, but I'll stick to your notation to avoid causing even more confusion.)
Analogously to the exact formula for $G(n,p)$, we can write a recursive formula for the probability that is an exact but useless answer:
$$
f(M,N) = \sum_{i=1}^M \binom{M-1}{i-1} \sum_{j=0}^N \binom{N}{j} \left(\frac{i^2}{M^2}\right)^j \left(\frac{(M-i)^2}{M^2}\right)^{N-j} f(i,j).
$$
The idea here is to sum over all ways to choose $i-1$ vertices to be in the same connected component as the first vertex, and then to choose $j$ edges to be in that component. Then the probability we get is the probability that those $j$ edges are between vertices in the connected component, that the other $N-j$ edges are not incident to those vertices, and that the connected component actually is connected.
We can, however, get very good approximations as $M \to \infty$. (These should be pretty good for $M$ that are not all that large.) The key turns out to be the number of isolated vertices. Setting $N = \frac12 M(\log M + C)$ for a new parameter $C$, we get that the probability that a given vertex is isolated is
$$
\left(1 - \frac{2M-2}{M^2}\right)^N \sim \exp\left(-\frac2M \cdot N\right) = \exp(-\log M - C) = \frac{e^{-C}}{M}.
$$
So the expected number of isolated vertices is $e^{-C}$. Though these events are not independent, they are very close to independent, and so the distribution of the number of isolated vertices is asymptotically Poisson with mean $e^{-C}$; therefore the probability tends to $e^{-e^{-C}}$ that there are no isolated vertices.
Meanwhile, more complicated connected components have vanished by the time $N$ is about $\frac12 M \log M$. For example, for a connected component of order $2$ you have to pay about the same price for all the missing edges (the expected number would be about $e^{-2C}$ if that's all that mattered) but also the edge connecting the two vertices you pick has to be present (for a penalty roughly on the order of $O(\log M/M)$). So $e^{-e^{-C}}$ is also the limiting probability that the graph is disconnected.
This means that when $N$ is much smaller than $\frac12 M \log M$, the graph is almost always disconnected; when $N$ is much larger, the graph is almost always connected.
A more detailed proof can be found as Theorem 4.1 here.
