If G = AB, then every element of G should be written as x=ab for a\in G and b\in G. for a\neq b since A\intersect B = empty. Consider the identity e of G. You cannot represent e as ab.
Is our assumption $A \cap B =\emptyset$ correct? or is $A \cap B = identity$?
If G = AB, we need to prove both inequalities. AB \in G and G \in AB
Obviously, AB is contained in G, since every x in AB is of the form ab which is also in G since G is a group.
Is G in AB. No. Since e \in G but e is not in AB by the assumption A \intersect B = empty. Hence G \neq AB
I hope this answers your question.