[If you wanted to--- it would take time, just because coming up with examples of anything involving matrices involves making a lot of choices--- you could find invertible matrices A, B, C, D satisfying ABCD = I but not ACD = B^(-1). This would show that isn't possible for any sequence of valid inferences to lead you from ABCD = I to ACD = B^(-1). But this is more complicated than solving your problem, and it doesn't solve your problem, so I'll skip it.]
I will now give one way of solving the problem.
[Theoretical point: it helps to note that from ABCD = I you know each of the matrices A, B, C, D must be invertible, since the appearance of a noninvertible matrix in any matrix product makes the entire product noninvertible. So knowing only that ABCD = I we can make free use of the existence of A^(-1), B^(-1), etc.]
From ABCD = I you can deduce that A^(-1) ABCD = A^(-1) I and hence that BCD = A^(-1) and hence that BCD A = A^(-1) A and hence that BCDA = I and hence that B^(-1) BCDA = B^(-1) I and hence that CDA = B^(-1).
Look over it a second time: all we are doing is a sequence of inferences like the type mentioned before: left multiplying both sides by A^(-1), right multiplying both sides by A, then left multiplying both sides by B^(-1). If you liked you could think of this as a single step: from ABCD = I you can deduce B^(-1) A^(-1) [ABCD] A = B^(-1) A^(-1) [I] A, and this simplifies to what you want. But even though our sequence of simple steps can be thought of as one complicated step, I think it is easier to solve equations these by thinking in terms of simple steps. I hope this helped.