$\newcommand\Om\Omega\newcommand\N{\Bbb N}\newcommand\R{\Bbb R}\newcommand\M{\mathscr M}\newcommand\A{\mathscr A}$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 $\mu_N(M)\ne\mu^*(M)$. $\quad\Box$


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