First, I note that the strong JFEP is the same as the medium JFEP, since two transitive sets such as $M[G]$ and $N[H]$ are isomorphic if and only if they are equal.
Next, note that the strong property is just too much.
Theorem. No consistent extension of ZF, which allows a model to be a forcing extension by adding a Cohen real, has the strong JFEP or the medium JFEP.
Proof. For any countable transitive model $M$ of ZF, there are $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common forcing extension. One can build $c$ and $d$ so that each of them separately is $M$-generic, but the combination $c\oplus d$ codes a real that collapses all of $M$. This is explained in my answer to this question. QED
This argument can be generalized beyond the forcing to add a Cohen real to any forcing that allows that coding argument to go through, and I believe this might include all forcing.
Theorem. There is a theory $T$ extending ZFC with the strong JFEP.
Proof. Assume without loss that there are transitive models of ZFC. Thus, there is a minimal transitive model. Let $T$ be the complete theory of the minimal transitive model $L_\alpha\models ZFC$. Now, the point is that if $M$ and $N$ are models of $T$, then actually $M=N$, since both must have the form $L_\beta$ for some $\beta$ but it cannot be that $\beta\gt\alpha$ and so both are equal to the minimal transitive model itself. Since $M=N$, of course it follows that they have a common forcing extension. QED
This idea seems to generalize to many other theories.
Meanwhile, it is easy to get the weak JFEP.
Theorem. Every complete theory trivially has the weak JFEP.
Proof. If $T$ is complete and $M,N\models T$, then they are already elementarily equivalent and they are forcing extensions of themselves by trivial forcing. QED