@OlaSande I am sorry this question got closed because there are still a couple more things that need to be said

@JeremyRickard Let $\Delta$ denote the set of elements of $G$ that have finitely many conjugates: this is a characteristic subgroup and is (locally finite)-by-abelian. Since our group $G$ in the question has finite cohomological dimension over $\mathbb Z$, it follows $\Delta$ is torsion-free abelian of finite rank. So some subgroup $H$ of finite index in $G$ centralises $\Delta$. Now your original line of reasoning works for $H$ so I think you successfully show $G$ must be virtually abelian. That is an interesting reduction.

Applying the Milnor sequence to the Baumslag-Solitar group mentioned in the question gives $H^3(G,M)\cong\lim^1H^2(BS(2,3),M)$ for any $G$-module $M$ where $G$ is the largest metabelian quotient of $BS(2,3)$ which is a valuable point of view but not quite the tree-action inspired view I suspect might be out there.

@GustavoGranja Very possibly. Please could you give a reference?

@JeremyRickard I'm sorry you deleted your answer because it looked plausible save for one point I noticed that the centre of the group ring can be a little larger than the group ring of the centre. Does this really cause a lot of difficulty or was there some other problem with your plan?

Can you give a reference for the noetherian commutative case? I can't even see why it's true for $R=\mathbb Z$ at this moment.

Let us continue this discussion in chat.

@IgorBelegradek Do you think my contibution is not really adding anything new? IF that is so, I perhaps should delete it.

Unfortunately I don't have Holt and Plesken's book to hand. But it sounds highly plausible that their argument, on the abstract algebra side, would be like mine.

@IgorBelegradek OK I have rewritten the originally flawed second paragraph and I think the new paragraph, while longer, is also simpler and may be correct this time.

The only ring homomorphsm $A\to \mathbb Z_2$ that does not preserve 1 is the zero homomorphism.

The Gallian and Buskirk calculation includes homomorphisms which do not preserve 1. In your case, only the zero homomorphism is multiplicative and does not preserve 1: you should exclude this case from your count.

Are you sure Gallian and Buskirk are looking at ring homomorphisms? Or are they counting arbitrary additive homomorphisms?

It would be helpful if you could include a definition of 'scattered' in your question. Also can you please point to a reference for the assertion that every k-space is a quotient of a disjoint union of compact Hausdorff spaces: is this a characterisation of k-spaces?

@OlaSande There is nothing confusing or contradictory here. Your question has been interpreted in two ways: if you are looking for a topological ring with additive structure $S^1$ this is not possible because compact rings are profinite and $S^1$ is not profinite. YCor's reply gives some indication how to prove this in the special case you are considering. If you ignore the natural topology on $S^1$ then there are compatible ring structures and there are many ways to construct them, for example see Goodwillie's comment. Non-unital rings can be called algebras but should not be called rings.

I guess any additive (topological) group can be a non-unital ring simply by defining the multiplication to be consistently zero.