We know that $\{I\}$ is a filter over $I$.
I'd like to show that $Th(M^I / F) = Th(M)$ as a consequence of Los's theorem.
Now, Los's theorem does not work in general for filters. See:
What can we say if we have a filter instead of an ultrafilter in Los's theorem?
Does it work for the filter $\{I\}$?
As long as $I$ has more than one element, the answer is no - consider what happens if we take the filtered power with respect to $\{I\}$ (which is just the usual Cartesian product) of a two-element structure, thinking in particular about the sentence $$\forall x,y,z(x=y\vee y=z\vee x=z).$$