I assume $m^*$ is an outer measure; and will use the fact that a set $E$ is measurable if and only if
$$m^*(E)=m^*(E\cap T)+m^*(E\cap T')$$ for all sets $T$.
Since $\cup_{k=1}^\infty E_k$ is measurable, your result is equivalent to proving that
$$\tag{1}
m^*(A)=\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr).
$$
Towards that end, note, by the the countable subadditivity of $m^*$
$$\eqalign{
m^*(A)
&\le m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr) \Bigr) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr)\cr
&\le\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr).
}
$$
With this in hand, we see that the desired result is trivial if $m^*(A)=\infty$.
Now assume $m^*(A)\ne\infty$.
To prove the reverse inequality needed,
we will prove by induction that for each $p\ge1$ and any set $A$
$$\tag{2}
m^*(A)=\sum_{k=1}^p m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^p E_k \Bigr)'\ \Bigr).
$$
Once this is done, since the sequence $\Bigl(m^*\Bigr(A\cap \bigl(\cup_{k=1}^p E_k \bigr)'\Bigr)\Bigr)_{p=1}^\infty$ is nonincreasing and bounded below by
$m^*\Bigr(A\cap \bigl(\cup_{k=1}^\infty E_k \bigr)' \Bigr)$,
it will follow that
$$
m^*(A)\ge\sum_{k=1}^\infty m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^\infty E_k\ \Bigr)'\Bigr),
$$
as desired.
So, on to the proof of $(2)$.
The base case follows from the fact that $E_1$ is measurable:
$$
m^*(A)=m^*(A\cap E_1)+m^*(E\cap E_1').
$$
Assume $(2)$ is true for $p=k$.
Since $E_{p+1}$ is measurable,
$$m^*(A)=m^*(A\cap E_{p+1})+ m^*(A\cap E_{p+1}')$$
Now, using $(2)$ applied to the set $A\cap E_{p+1}'$,
$$
\eqalign{
m^*(A)&=m^*(A\cap E_{p+1})+ m^*(A\cap E_{p+1}')\cr
&=m^*(A\cap E_{p+1})+\sum_{k=1}^p m^*(A\cap E_{p+1}' \cap E_k) +m^*\Bigl( A\cap E_{p+1}'\cap \Bigl( \bigcup_{k=1}^p E_k\ \Bigr)'\Bigr)\cr
&=m^*(A\cap E_{p+1})+\sum_{k=1}^p m^*(A\cap E_k) +m^*\Bigl( A\cap E_{p+1}'\cap \Bigl(\ \bigcup_{k=1}^p E_k\ \Bigr)'\Bigr)\cr
&=\sum_{k=1}^{p+1} m^*(E_k\cap A) +m^*\Bigr(A\cap \Bigl(\ \bigcup_{k=1}^{p+1} E_k\ \Bigr)'\Bigr);
}
$$
where, in the third equality, we used the fact that ${E_k\subset E_{p+1}'}$. This is exactly what we need to finish the proof by induction. $(2)$ holds for all $p$.
