First note that $M$ is countable while $M^*$ has size continuum. So the two structures are never isomorphic. But you may instead ask when/if they are elementarily equivalent; and the answer is “sometimes”.

Example 1: Let $\mathcal{C}$ be the class of finite graphs. Then $M$ is the countable random graph, and one can have that $M^*$ is elementarily equivalent to $M$ (for example by taking an ultraproduct of Paley graphs).

Example 2: Let $\mathcal{C}$ be the class of finite linear orders. Then $M$ is isomorphic to $(\mathbb{Q},<)$, while $M^*$ is a ($\aleph_1$-saturated) discrete linear order with endpoints.

EDIT: In the question “an ultraproduct of members of $\mathcal{C}$” is ambiguous, since perhaps you allow one to take an ultraproduct of a subclass. This doesn’t make much difference for Example 2, but in Example 1 one could potentially get different answers for $M^*$ by varying the subclass (or even the ultrafilter). For example, an ultraproduct of the class of triangle-free graphs will not be a model of the theory of the random graph.