Let's start a bit simpler than your question: assume A is a|xx and B is b|yyy
Thus A represents the set { a, xx } and B, the set { b, yyy }.
Then A ⊗ B represents { (a, b), (a, yyy), (xx, b), (xx, yyy) }.
And thus (A ⊗ B)+ is { (a, b), (a, yyy), (xx, b), (xx, yyy), (aa, bb), (aa, byy), (axx, bb), (axx, byyy), (axx, byyy), (axx, yyyyyy), (xxxx, byyy), (aaa,bbb), .... }
all the possible concatenations of elements of A ⊗ B,
with the trivial mapping of "sequences of couples" to "couples of sequences" : (i,u)(j,v) corresponds to (ij, uv).
Formally (A ⊗ B)+ is the set { (r^n, s^n) } with the SAME n, for all possible r in A and s in B (and for all n >= 1).
With the same above example, A+ is { a, xx, aa, axx, xxxx, aaa, aaxx, ...}
and B+ is { b, yyy, bb, byyy, yyyb, yyyyyy, bbb, bbyyy, ....}
and A + ⊗B+ is the cross product of these two sets : { (a,b), (a,yyy), (a,bb), (a, byyy), (a, yyyb), (a, yyyyyy), (a, bbb), (a, bbyyy), (a, ....), ..., (xx, b), (xx, yyy), ... }
Formally, A + ⊗B+ is the set { (r^n, s^m) } for all possible r in A and s in B (and for all n >= 1 and m >= 1), notice that here m is not necessarily equal to n.
Now, in your question, you included "+" in A and B already. This does not change the answer but makes it more complex to write and understand I guess since A and B are no longer finite as in the above simpler example. But the formal answer is the same.
Makes sense ?