Let's simplify the system. A large block (mass $M$) ejects small blocks (mass $m$) using springs (so instantaneous $\Delta v$). Let's consider all velocities in the Lab frame (Newtonian physics will do fine for this).
We will work out the energies in the two cases (both blocks in same direction versus blocks in opposite directions) and see if the energy is the same or different.
Case 1: Opposite directions
We eject the first small block and it moves off to the right with velocity $v_1$. The big block (with a second small block still attached, so mass $= M + m$) recoils to the left with velocity $v_0$.
Now we eject the second small block to the left. It shoots off with velocity $v_2 = v_2' + v_0$, where $v_2'$ is its velocity relative to the block. Note that $v_2' \ne v_1$ because the mass of the big block is now $M$ so it's easier to stop (we had to reduce the spring's force). The big block stops.
We now have a stationary big block and two small blocks moving off in opposite directions. 512
However... the two small blocks do not have the same velocity! (The original question assumed a symmetry that, in fact, is not there).
If we calculate the energy, we now have:
$E_1 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 $
$= \frac{1}{2}m(v_1^2 + v_2^2)$
$= \frac{1}{2}m(v_1^2 + (v_0 + v_2')^2)$
$= \frac{1}{2}m(v_1^2 + v_0^2 + 2v_0v_2' + v_2'^2)$
Case 2: Same direction
Consider now what happens when the blocks are shot off in the same direction. The first block shoots off as before at $v_1$. The big block recoils at $v_0$.
Now we shoot off the second block. The second block now has a reduced velocity in the lab frame, because the big block is moving in the opposite direction, so $v_2 = v_2' - v_0$.
Now the energy calculation is:
$E_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \frac{1}{2}M(2v_0)^2$
$= \frac{1}{2}m(v_1^2 + v_0^2 - 2v_0v_2' + v_2'^2) + 2Mv_0^2$
Now we equate the $E_1$ and $E_2$ case equations and see where that leads us. If it leads to something absurd, we know the energies are not equal (as the OP suspects).
$\frac{1}{2}m(v_1^2 + v_0^2 + 2v_0v_2' + v_2'^2) = \frac{1}{2}m(v_1^2 + v_0^2 - 2v_0v_2' + v_2'^2) + 2Mv_0^2$
$\therefore mv_0v_2' = -mv_0v_2' + 2Mv_0^2$
$\therefore mv_2' = Mv_0$
This is simply the equation of conservation of momentum for the 2nd small block shooting off the big block. Since we know this must be true, if must be that the energies in the two cases are actually the same.
I think the OP's mistake was to assume that the small blocks (in his case, the rocket exhausts) had the same momentum (and energy) in both cases. They don't. 512