A more general question is as follows. Let $\mu$ and $\nu$ be two measures on a measurable space $(S,\Sigma)$. Does it then always follow that $(\mu+\nu)^*=\mu^*+\nu^*$, where ${}^*$ denotes the corresponding outer measure?

The answer to this question is yes. Indeed, for any $E\subseteq S$, let $\mathcal{A}_E$ denote the set of all sequences $(A_j)$ of pairwise disjoint members of $\Sigma$ such that $\bigcup_j A_j\supseteq E$. Then
\begin{equation}
\mu^*(E)+\nu^*(E)=\inf_{(A_j)\in\mathcal{A}_E}\sum_j\mu(A_j)
+\inf_{(A_j)\in\mathcal{A}_E}\sum_j\nu(A_j)
\end{equation}
\begin{equation}
\le\inf_{(A_j)\in\mathcal{A}_E}\Big(\sum_j\mu(A_j)+\sum_j\nu(A_j)\Big)
=\inf_{(A_j)\in\mathcal{A}_E}\sum_j(\mu+\nu)(A_j)
= (\mu+\nu)^*(E).
\end{equation}

To prove the reverse inequality, $(\mu+\nu)^*(E)\le\mu^*(E)+\nu^*(E)$, take any real numbers $a$ and $b$ (if they exist) such that $\mu^*(E)<a$ and $\nu^*(E)<b$. Then for some sequences $(A_i)$ and $(B_j)$ in $\mathcal{A}_E$ one has $\sum_i\mu(A_i)<a$ and $\sum_j\mu(B_j)<b$. Let $C_{i,j}:=A_i\cap B_j$. Somehow enumerating the pairs $(i,j)$, we may assume that the double-sequence $(C_{i,j})$ is in $\mathcal{A}_E$. So,
\begin{equation}
(\mu+\nu)^*(E)\le\sum_{i,j}(\mu+\nu)(C_{i,j})
=\sum_i\sum_j\mu(C_{i,j})+\sum_j\sum_i\nu(C_{i,j})
\end{equation}
\begin{equation}
=\sum_i\mu(A_i)+\sum_j\nu(B_j)<a+b,
\end{equation}
for any real $a$ and $b$ such that $\mu^*(E)<a$ and $\nu^*(E)<b$.
Hence, $(\mu+\nu)^*(E)\le\mu^*(E)+\nu^*(E)$. QED