Prove or disprove - graph doesn't exists
If the question is - "Does there exists a directed graph $G=(V,E)$ (whose undirected version is connected) such that each DFS run on $G$ (preferably with different choices of root) always return exactly two trees?", then the answer is No.

Compress all SCCs into supernode, and let $G^*=(V^*, E^*)$ be resulting graph with $V^*=\{S_1,S_2,\ldots, S_k\}$. Each SCC ($S_i$'s) must have either an incoming edge or an outgoing edge. Now,


At least one SCC must have a zero incoming edge (say it is $S_1$).
At least one SCC must have a zero outgoing edge (say it is $S_k$).


If you take a root in SCC $S_k$ as compared to $S_1$, then DFS will give one more tree for sure.