Recall that $\mu (A)=\mu^{*}(A)$ if $A \in \mathcal A$ and that $\mu^{*}$ is monotone.
Whenever $B_i,C_i \in \mathcal A$, $B_i's$ are disjoint, $\bigcup B_i \subset A$ and $A \subset \bigcup C_i$ we have $\sum\limits_{k=1}^{n} \mu(B_i) =\mu (\bigcup_{k=1}^{n} B_i) =\mu^{*} (\bigcup_{k=1}^{n} B_i)\leq \mu^{*} (\bigcup_{k=1}^{\infty} C_i)\leq \sum\limits_{k=1}^{\infty} \mu(C_i) $ for each $n$. Letting $n \to \infty$ we get $\sum\limits_{k=1}^{\infty} \mu(B_i) \leq \sum\limits_{k=1}^{\infty} \mu(C_i) $. The proof is now finished by taking infimum over all $(C_i)$ and supremum over all $(B_i)$.
Now that the measures are assumed finite I think this is doable. First we wish to show that $\mu_*(A) = \mu(\Omega) - \mu^*(A)$. Let $\epsilon > 0$. Choose $B_i$ so that $\cup_i B_i \subset A$, and $\mu_*(A) \le \mu(\cup_i B_i) + \epsilon$. It follows since $\mu$ is additive that for $C\in \mathcal{A}$, $\mu(C) = \mu(\Omega) - \mu(C^c)$. A picky detail, but additionally there exists $n$ so that $\mu(\cup_i B_i) \le \mu(\cup_{i=1}^n B_i) + \epsilon$. This is needed since $\cup_{i=1}^n B_i \in \mathcal{A}$ ($\mathcal{A}$ is an algebra). With this, and noting that $ A^c \subset \left(\cup_{i=1}^n B_i \right)^c $,
$$
\mu_*(A) \le \mu(\cup_i B_i) + \epsilon \le \mu(\cup_{i=1}^n B_i) + 2\epsilon = \mu(\Omega) - \mu(\left(\cup_{i=1}^n B_i \right)^c) + 2\epsilon \le \mu(\Omega) - \mu^*(A^c) + 2\epsilon.
$$
Starting with a cover $C_i$ such that $A^c \subset \cup_i C_i$ and $\mu(\cup_i C_i) \le \mu^*(A^c) + \epsilon$, one can show very similarly that
$$
\mu^*(A^c) \ge \mu(\Omega) - \mu_*(A) -2\epsilon.
$$
Manipulating a bit and putting the above equations together we have
$$
\mu(\Omega)-\mu^*(A^c) - 2\epsilon \le \mu_*(A) \le \mu(\Omega)-\mu^*(A^c) + 2\epsilon.
$$
Since $\epsilon$ is arbitrary, we have $\mu_*(A) = \mu(\Omega)-\mu^*(A^c)$. Now this implies that the collection of sets $\mathcal{B}$ is closed under the complement. Suppose $A\in \mathcal{B}$ so that $\mu_*(A)= \mu^*(A)$. Then $\mu_*(A^c) =\mu(\Omega)-\mu^*(A)= \mu(\Omega)-\mu_*(A)= \mu^*(A^c)$.
It also came up in the comments to show that $\mathcal{B}$ is closed under intersection. To that end let $\epsilon>0$, and suppose $A_1,A_2 \in \mathcal{B}$. As a result we may choose $B_{i,1},B_{i,2},C_{i,2},C_{i,2}$ with the following properties: 1) $\cup_i B_{i,j} \subset A_j \subset \cup_i C_{i,j}$, $j=1,2$. 2) $\mu( \cup_{i} C_{i,j} / \cup_{i} B_{i,j}) < \epsilon/2$, $j=1,2$. Evidently then $(\cup_i B_{i,1}) \cap (\cup_i B_{i,2}) \subset A_1 \cap A_2 \subset (\cup_i C_{i,1}) \cap (\cup_i C_{i,2})$. Further, it can be shown that $(\cup_{i } C_{i,1} \cap C_{i,2}) /(\cup_{i } B_{i,1} \cap B_{i,2}) \subset (\cup_{i} C_{i,1} / \cup_{i} B_{i,1}) \cup (\cup_{i} C_{i,2} / \cup_{i} B_{i,2}). $ It follows that $\mu ( (\cup_i C_{i,1} \cap \cup_i C_{i,2}) /(\cup_i B_{i,1}) \cap (\cup_i B_{i,2})) < \epsilon$, giving that $\mu^*(A_1 \cap A_2) - \mu_*(A_1\cap A_2) < \epsilon$. Since $\epsilon$ is arbitrary, $\mu^*(A_1 \cap A_2)=\mu_*(A_1\cap A_2)$.
Best Answer
Example when $E$ is closed:
Let $E = [1,1+1/2] \cup [2,2+1/4] \cup [3,3+1/8] \cup \cdots = \bigcup_{k=1}^\infty [k,k+2^{-k}]$.
Then $m(E) = 1$, but $m(O_m) = \infty$ for all $m$.
Example when $E$ is open:
Let $\{r_k\}$ denote an enumeration of the rational numbers in $(0,1)$.
Fix $\epsilon > 0$ and for each $k$ choose $\delta_k$ so that the interval $I_k = (r_k - \delta_k,r_k+\delta_k)$ satifies
One way to do this is to take $\delta_k = \min\{ \epsilon/2^{k+1}, \mathrm{dist}(x,\{0,1\}))$.
Let $E = \cup_k I_k$. Then $E$ is open and $m(E) \le \sum \ell(I_k) < \epsilon$.
On the other hand, since $\{r_k\} \subset E$ each set $O_m$ contains $[0,1]$ so that $m(O_m) \ge 1$ for all $m$.