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