Let G be a bipartite graph that does not have a perfect matching, but if we add any new edge to G, then the new graph G+ is no longer bipartite, but has a perfect matching. What can G be?
I found one such graph, for example, 3 vertices on the left part connected with an edge to a single vertex on the right part of the graph, new edge can be created only on the left part and that will yield perfect matching and G will no longer be a bipartite graph. This is clearly not possible with a bipartite graph with equal number of vertices on both sides as it says any new edge.The new edge must be added on either part of the graph so that it is no longer bipartite.
Any idea what kind of graph G can be or is there any other such G?
If there is a vertex on the left not connected to a vertex on the right, we could add an edge joining them and the graph would still be bipartite. So all vertices on the left must be connected to all vertices on the right: you have a complete "bipartite graph". $K(m,n)$ is the complete bipartite graph with $m$ and $n$ vertices in the two parts. In your case you need $m \ne n$, so there is no perfect matching.
Now if you have $K(m,n)$ and join two vertices on one side, it's no longer bipartite. Remove the two vertices you have joined and all the edges incident on them, and you are left with $K(m-2,n)$ if those two vertices were on the left, or $K(m,n-2)$ if those two vertices were on the right. In order for the new graph to have a perfect matching, you must have $m-2 = n$ or $m = n-2$ respectively. Thus the graph you started out with must have been $K(n,n+2)$ or $K(n+2,n)$ for some $n$.
But it says "any new edge": if you added a new edge on the side that had fewer vertices, it would make things worse. The only way this can work is that it is impossible to add a new edge on that side, which only happens if that side only had $1$ vertex. Thus the original graph must have been $K(3,1)$ (or $K(1,3)$ which is the same thing).