Yes, they are equivalent generalizations of the construction of Lebesgue Measurable Sets on $\mathbb{R}^n.$ Throughout, let $d$ be the pseudometric induced by $\mu^* .$
$\textbf{ Part 1) }$ [$\mathscr{A}_{\mu } \subset \mathscr{M}_{\mu^*}$ ]
By countable subaddivity, for measurable $A$ and all $E \subset X$ we have $\mu^* (E ) \le \mu^* (A \cap E ) + \mu^* (A^c \cap E),$ so that we need only show the reverse inequality.
When $A \in \mathscr{A}$ , we have for any approximate outer measure $l$ of $E$ that there exists a countable $\{ E_n \}$ covering $E$ such that
$l \ge \displaystyle\sum_{n=1}^{\infty } \mu (E_n) =$ $\displaystyle\sum_{n=1}^{\infty } [ \mu (E_n \cap A) + \mu (E_n \cap A^c)] = $
$\displaystyle\sum_{n=1}^{\infty } \mu (E_n \cap A) + \displaystyle\sum_{n=1}^{\infty } \mu (E_n \cap A^c ) =$ $\mu (E \cap A) + \mu (E \cap A^c )$
Now for $A \in \mathscr{A}_{\mu },$ consider the same notation as before. Note that if $l$ is an approximate outer measure for $E$ then we can write $ l = \displaystyle\sum_{n=1}^{\infty } l_n + g_n , $ where each $l_n$ is an approximate outer measure for $E_n \cap A_n ,$ and $g_n$ for $E_n \cap A^c.$ Hence
$l \ge \displaystyle\sum_{n=1}^{\infty} [\displaystyle\sum_{k=1}^{\infty } \mu (A_k \cap E_n) ] + [\displaystyle\sum_{k=1}^{\infty } \mu (B_k \cap E_n) ] = \mu (E \cap A) + \mu (E \cap A^c)$
$\textbf{ Part 2) }$ [$\mathscr{M}_{\mu^* } \subset \mathscr{A}_{\mu}$ ]
Suppose $\mu^* (A) $ is finite and that $A$ satisfies the Caratheodory criterion. Then for any $\epsilon >0,$ there exists $E \in \mathscr{A_{\mu }} $ such that $\mu^* (E) - \mu^* (A) \le \epsilon .$ Hence
$d(E, A) = \mu^* ( (A-E) \cup (E-A) ) =$ $\mu^* (E-A ) =$ $ \mu^* (A^c \cap E) =$
$ \mu^* (E) - \mu^* (A \cap E) = $ $\mu^* (E) - \mu^* (A) \le \epsilon $