Relation between random graph models $G^{(B)}_{n,m}$ and $G_{n,m}$

In Frieze, Alan; Karoński, Michał, Introduction to random graphs, in Section 1.3 Pseudo-Graphs, there is a model of random multi-graphs, which is denoted as $$\mathbb{G}^{(B)}_{n,m}$$.

Def. A random multi-graph $$\mathbb{G}^{(B)}_{n,m}$$ contains $$m$$ edges, where every edge is chosen uniformly at random and independently from $${[n] \choose 2}$$.

It is also shown in the same section that $$\mathbb{G}^{(B)}_{n,m}$$ and $$\mathbb{G}_{n,m}$$ are equivalent, when $$m = O(n)$$. ($$\mathbb{G}_{n,m}$$ is the simple random graph with $$m$$ edges.) The equivalence is in the sense that for a graph property $$\mathcal{P}$$, if $$\mathbb{P}(\mathbb{G}^{(B)}_{n,m} \in \mathcal{P}) = o(1)$$, then $$\mathbb{P}(\mathbb{G}_{n,m} \in \mathcal{P}) = o(1)$$, for $$m = O(n)$$.

My question is what happens in the case when $$m = \omega(n)$$? For example, $$\mathbb{G}_{n,m}$$ is connected a.a.s. when $$m = \frac{1}{2}(n\log{n} + \omega(1))$$. What about $$\mathbb{G}^{(B)}_{n,m}$$?

The relation between $$\mathbb{G}_{n,m}$$ and $$\mathbb{G}^{(B)}_{n,m}$$ is not quite an equivalence. All we have is that if the random multigraph does not have a property almost surely, then the random simple graph also does not have this property, within the range $$m=O(n)$$. The reason essentially is that is within this range, $$m$$ is so small that there is a constant probability that the random multigraph turns out to be a simple graph. However, the converse implication need not hold, for example for the graph property "The graph is not simple" will hold with probability $$0$$ for the simple random graph, but not necessarily for the random multigraph.
For larger values of $$m$$, $$\mathbb{G}^{(B)}_{n,m}$$ will become non-simple, so there wouldn't be any natural implications anymore. For the specific question you ask, I think Theorem 4.2 from the same book ($$\mathbb{G}_{n,m}$$ becomes connected precisely when no isolated vertices remain) can give you an answer, provided that the same result holds for $$\mathbb{G}_{n,m}^{(B)}$$. I haven't read through the entire proof carefully to check this, but I would be very surprised if it did not generalize for some reason.