Thanks -- it was a very nice answer! I've been thinking about the natural sum and product for the last half hour, and I'm convinced that van der Waerden's Theorem holds in that case. I'm writing this now in a comment because I won't have time to write out a proof until tomorrow.

@Joel: If we use $\alpha + F \cdot \beta$ then choosing the right $\beta$ will make $F \cdot \beta$ a single point, which makes the conclusion trivial. Using the natural sum and product could be very interesting, though (I hadn't thought of that). I think (a modification of) the ultrafilter proof of van der Waerden's Theorem might go through in that case, but I'll need to double-check the details before I can say for sure. If it works out I'll let you know.

If $L$ is a minimal left ideal of $(\omega^*,+)$ and if $\mathcal F$ is the Stone dual of $L$, then $\mathcal F$ has the property you describe. $\mathcal F$ is neither an ultrafilter nor the cofinite filter, but it can be sent to an ultrafilter under a finite-to-one map. So I don't know if this is really the kind of answer you're looking for. If you want to see the details, let me know and I'll be happy to write them out.

@Ben, concerning your question about $\omega^\omega$, it looks like Joel's answer below applies equally well to $\omega^\omega$, or for that matter any indecomposable ordinal. So $\omega$ is the unique ordinal satisfying van der Waerden's Theorem.

Yes, you're right -- my bad. I'll edit accordingly.

I don't, but I think it's another good question. It's fairly easy to show that if it holds for some ordinal $\alpha$ then $\alpha$ must be indecomposable (a power of $\omega$).