$(B \otimes C) \otimes (D \otimes E)$ is isomorphic to $B \otimes C \otimes D \otimes E$ Let $B, C, D, E$ be $A$-modules. Is there a way to show that
$(B \otimes C) \otimes (D \otimes E)$ is isomorphic to 
$B \otimes C \otimes D \otimes E$ using the result that
$(M \otimes N) \otimes P$ is isomorphic to $M \otimes N \otimes P$
for any $A$-modules $M, N$ and $P$ ?
 A: To answer your direct question, yes, there is certainly a way to prove this, becuase it's certainly true. How to do it? Anytime tensor products show up, your best bet is appealing to the universal property, because it's the only thing you really know about the tensor product (especially if you're doing this over an arbitrary commutative ring $A$).
I don't think using the result you mention will help, because it doesn't say anything about the quaternary product. But if you have the proof of the ternary result, the proof you want (for the quaternary case) will be pretty much identical: use universal properties to construct unique homomorphisms in each direction, and then show (trivially) that they are inverses of each other.
Edit: Obviously it has been too long since I did any tensor algebra myself. The real way to prove this is just to show that $(B \otimes C) \otimes (D \otimes E)$ satisfies the quatrilinear universal property (by appealing to the various bilinear universal properties involved), and must therefore be isomorphic to the quaternary tensor product, because the tensor product is unique up to isomorphism. The end! (No mucking about with trying to show inverses or anything like that.)
A: I will assume that you already know that
$$
  (M \otimes N) \otimes P \cong M \otimes (N \otimes P).
$$
Then, I will show that "whatever order" the parenthesis appear, the result
of calculating $X_1 \otimes X_2 \otimes \dotsb \otimes X_n$
(parenthesis were ommited since I did not have enough creativity to create
a notation for it) is always the same as associating from left to right.
That is, it is always isomorphic to
$((((X_1 \otimes X_2) \otimes X_3) \dotsb) \otimes X_n)$.
We use induction on $n$, and we know that the statement is true for $n = 3$.
Take an "innermost term". That is, some $(X_k \otimes X_{k+1})$, and treat it
as a single term. Then, the inductive hypothesis implies that the product is
isomorphic to
$$
  ((((X_1 \otimes X_2) \otimes X_3) \dotsb \otimes (X_k \otimes X_{k+1})) \otimes X_n).
$$
Case 1 ($k = 1$): Nothing to do.
Case 2 ($k > 2$): From the equation above,
we can take $k = 1$ and reduce this to Case 1.
Case 3 ($k = 2$): The equation becomes
$$
  ((((X_1 \otimes (X_2 \otimes X_3)) \dotsb ) \otimes X_n).
$$
Now, just use the case for $n = 3$, that is,
$X_1 \otimes (X_2 \otimes X_3) \cong (X_1 \otimes X_2) \otimes X_3$,
to get to the desired conclusion.
