Well, you forgot to mention that pieces affected by A != pieces affected by C. That's very important to remember. In fact, that very fact can be used as a generalized process of doing this. I made an effort to come up with a different example where B != D (since that appeared to be too trivial).you gave one set of conditions (B = D etc) under which merging is possible, as [A, B][C, B] = [AC, B].
With [A, B] = [R U2 R',F] and [C, D] = [[B' D': M' D L' D' M D L D'], S' R2 S R2],
both C and D are "independent" from A, B and [A, B].
That is, [R U2 R',F] [[B' D': M' D L' D' M D L D'], S' R2 S R2] can be merged to
[F: [F' [B' D': M' D L' D' M D L D'] , S' R2 S R2 R U2 R']]
(Notice that I conjugated once to make this work, but this is still a commutator nonetheless.)
Does this example show things better for you? If you were to ask me when two commutators cannot be merged, if their pieces are not independent (whether both pieces from one commutator are from the other commutator--this example--or if one piece from each is independent--first example), one could merely attempt to combine any two commutators that do not follow the restrictions I gave. To me, that's proof that those particular two commutators cannot be merged into one (well, not to one that yields the same permutation as the product of the two commutators).
Why? If no pieces from the commutators are independent, then one will conflict with the other, at some level or another. You can still possibly merge them into a commutator, but that commutator will not have the exact same effect on the cube as their product due to the moves from one commutator affecting pieces the other affects.
I still don't see why when two commutators cannot merge doesn't mean one of them cannot merge with another that does the same permutation of the other, as long as the new commutator being used follows the restrictions I illustrated here.
If I am still not understanding what is being asked, I apologize. I mainly wanted to show another example which elaborated my method more.