Without assuming that $A$ and $B$ are disjoint, this is wrong for silly reasons, as was pointed out in the comments: if $A$ has positive (outer) measure and $B$ overlaps with $A$ in a non-trivial way, we always have $\mu^\ast (A \cup B) \lt \mu^\ast (A) + \mu^\ast(B)$. For a specific example, we may take $A = [0,2]$ and $B=[1,2]$ so that $2 = \mu^\ast(A \cup B) \lt 3 = \mu^\ast[0,2]+\mu^\ast[1,2]$ with respect to Lebesgue outer measure.
However, if $A$ is measurable and $B$ is disjoint from it, we can take $C = A \cup B$ and $E = A$ and use Carathéodory's criterion (the definition of measurability you gave) together with $C \cap A = A$ and $C \cap A^c = B$ to see that
\mu^\ast (A \cup B) = \mu^\ast(C) = \mu^\ast (C \cap A) + \mu^\ast(C \cap A^c) = \mu^\ast(A) + \mu^\ast(B),
which is the desired equality.