Fundamental groups of codimension 1 manifold complements

I think you should be able to show this using Scott's Core theorem. Pick a compact core $C$ in $M$. Now you might want to distinguish some cases, but you should get somthign like the generation of $\pi_1M_i$ by the image of $\pi_1(C\cap M_i)$ and $\pi_1S$. You should also assume properly embeddedness for this to work.

EDIT: So as pointed out in the comments my answer was just too sloppy to understand so I decided to give a proof.

First: how to you come up with the idea of the proof? So my attempt was to look at graphs of groups or groupoids to avoid issues related to base point, connectedness and so on. And now you just start to look at suitable sets covering the core $C$ and the seperating surface $S$ and partition them more and more (and sometimes less) to get nice generating sets for different parts. You get a graph of groups and which contains the relations among the fundamental groups of the sets. Eventually I got multiple partitions of sets which would do the trick. Here is a particularly easy one, to which I arrived by using the above technique, but which actually only consists of two sets.

So let* $K= C\cup \nu S$ and $K_i = K \cap M_i$, where $\nu S$ is a trivial tubular neighborhood (trivial by seperating property and $M$ is connected) small enough that everything looks like a product there (especially transverse intersections with $C$). Let $k^i:S \hookrightarrow K_i$ and $l^i$ the same map post composed to embed $S$ into $M_i$. Now we see that by Van Kampen (here we are again slightly ignoring connectedness of the $K_1,K_2$**) we have $$ \pi_1K = \pi_1K_1 *_{k_*\pi_1S} \pi_1K_2. $$

So this is the pushout $(\pi_1 k^1,\pi_1 k^2)$ and we have a map from it $$c:\pi_1K \twoheadrightarrow \pi_1M = \pi_1 M_1 *_{l_*\pi_1S} \pi_1M_2.$$ But we have more structure to that pushout map, namely that it arises not only from a map $\pi_1 K_i \to \pi_1M$ but that those maps each factor through $\pi_1M_i$. Hence our maps $\pi_1 K_i \to \pi_1M_i$ generate at least the part $\pi_1M_i - l^i_*\pi_1S$. That is because $c_*\pi_1K = c_*\pi_1K_1 \cup c_*\pi_1K_2$ and these two images (or rather the two factors) are always disjoint except for the image of the intersection. But as we have the surface in there from the beginning (algebraically you could also say that we are amalgamating over a larger subgroup already which is preserved, i.e. onto) that is no problem and we see that we get surjections $$\pi_1K_i \twoheadrightarrow\pi_1M_i.$$

So we showed: picking a modified core for $M_i$ gives us a finitely generated fundamental group.

* You might want to check that this proof won't work with $C$ and $C_i$ and that this is not even true in general.

** So by choosing our $K$ we made sure, that every component of $C-(C\cap S)$ touches $\cap S$, so we have that $S$ cuts this component containing $S$ into two components. So you could only look at the connected part and do the other components seperately eventually. This is why I ignore this issue. And another argument might be to just choose $C$ connected, which makes this obsolete as well.

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Manifolds

Differential Topology

Algebraic Topology

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