$\newcommand\Om\Omega\newcommand\N{\Bbb N}\newcommand\R{\Bbb R}\newcommand\M{\mathscr M}\newcommand\A{\mathscr A}$The equality $\Om_N=\Om_C$ follows immediately from Theorem 1.4 and Remark 1.5 in [this paper][1] or [its preprint version][2], where a more general setting is considered -- without requiring what is called the premeasure in the OP to be finite. In that paper, $\M,\M_{\mathsf{Ca}},\A,m,m^*$ correspond to $\Om_N,\Om_C,\Om_0,\mu_0,\mu^*$ in the OP, respectively. (Also, in that paper, the extension of $m$ from $\A$ to $\M$ is defined simply as the restriction of $m^*$ to $\M$.)

However, in general $\mu_N\ne\mu^*|_{\Om_N}$. E.g., let $X=\R$, let $\Om_0$ be the powerset of $\N$, and let $\mu_0$ be any finite measure on $\Om_0$. Then $\Om_N$ is the powerset of $\R$ and for any $M\in\Om_N$ such that $M\not\subseteq\N$ we have $\mu_N(M)\le\mu_N(\N)<\infty=\mu^*(M)$ (because no union of members of $\Om_0$ contains such a set $M$), so that $\mu_N(M)\ne\mu^*(M)$. $\quad\Box$ 

---

On the other hand, if $\mu_0$ is $\sigma$-finite, so that $X=\bigcup_{k=1}^\infty A_k$ for some $A_k$'s in $\Om_0$, then $\mu_N=\mu^*|_{\Om_N}$, by the uniqueness of the measure extension -- see e.g. Theorem 1.3 in the mentioned paper. 

Moreover, one has the following characterization of the condition $\mu_N=\mu^*|_{\Om_N}$: 

>**Proposition 1:** The following three conditions are equivalent to one another: 
> 1. $\mu_N=\mu^*|_{\Om_N}$.
>  2. For any $M\in\Om_N$ such that $\mu_N(M)=0$ there exist $U_1,U_2,\dots$ in $\Om_0$ such that $M\subseteq\bigcup_{k=1}^\infty U_k$.
> 3. For any $M\in\Om_N$ such that $\mu_N(M\cap A)=0$ for all $A\in\Om_0$ there exist
    $U_1,U_2,\dots$ in $\Om_0$ such that $M\subseteq\bigcup_{k=1}^\infty
    U_k$.

*Proof:* That conditions 2 and 3 are equivalent to each other follows immediately from the definition of $\mu_N$. 

That condition 1 implies condition 2 follows immediately from the definition of $\mu^*$. Indeed, if this implication were false, then for some $M\in\Om_N$ we would have $\mu_N(M)=0$ and $\mu^*(M)=\infty$. 

It remains to show that condition 2 implies condition 1. Assume that condition 2 indeed holds. Take any $M\in\Om_N$. We have to show that $\mu_N(M)=\mu^*(M)$. First here, note that 
$$\mu_N(M)=\sup_{A\in\Om_0}\mu_N(M\cap A)
=\sup_{A\in\Om_0}\mu^*(M\cap A)\le\mu^*(M).$$

It remains to show that $\mu^*(M)\le\mu_N(M)$. Here without loss of generality $\mu_N(M)<\infty$. By the definition of $\mu_N$, there is a sequence $(A_k)_{k=1}^\infty$ in $\Om_0$ such that $\mu_N(M\cap A_k)\to\mu_N(M)$. So, $\mu_N(M\cap A)=\mu_N(M)$, where $$A:=\bigcup_{k=1}^\infty A_k=\bigcup_{k=1}^\infty B_k\in\Om_N$$ 
and $B_k:=A_k\setminus\bigcup_{j=1}^{k-1}A_j\in\Om_0$, so that the $B_k$'s are pairwise disjoint. Therefore and because $\mu_N(M)<\infty$, we have $\mu_N(R)=0$ for $R:=M\setminus A$. So, by condition 2,  there exist $U_1,U_2,\dots$ in $\Om_0$ such that $R\subseteq\bigcup_{k=1}^\infty U_k$. So, 
$$\mu^*(R)=\mu^*\Big(\bigcup_{k=1}^\infty (R\cap U_k)\Big)
\le\sum_{k=1}^\infty\mu^*(R\cap U_k)
=\sum_{k=1}^\infty\mu_N(R\cap U_k)=0,$$
since $\mu_N(R)=0$. So, $\mu^*(R)=0$ and hence 
$$\mu^*(M)=\mu^*((M\cap A)\cup R)\le\mu^*(M\cap A)+\mu^*(R)
=\mu^*(M\cap A)
=\mu^*\Big(\bigcup_{k=1}^\infty(M\cap B_k)\Big)
\le\sum_{k=1}^\infty\mu^*(M\cap B_k)
=\sum_{k=1}^\infty\mu_N(M\cap B_k)
=\mu_N(M\cap A)\le\mu_N(M),$$
so that $\mu^*(M)\le\mu_N(M)$. $\quad\Box$

  [1]: https://link.springer.com/article/10.1007/s11117-017-0507-8
  [2]: https://arxiv.org/abs/1702.01142