Can the Kan-Thurston theorem be turned into some kind of equivalence between groups and spaces? I not really familiar with these subjects. I read  this question and I was really surprised by the answer. My question is probably vague (so please do bear with me). 
The cited question/answer suggests that any integral homology of a simply connected space can be realized as integral homology of some Eilenberg-MacLane space. Does it mean that connected spaces are the same thing (up to ?) as groups in some sense ? My question is vague, very vague. I will be happy to see a correct formulation of my question (if it makes sense). 
 A: Here's something that might be considered an answer.
In Section 11 of Baumslag, Dyer, Heller, "The topology of discrete groups", the following theorem is proved, basically building on the ideas of the Kan-Thurston theorem.  
A perfect homomorphism of groups $G\to H$ is a surjective group homomorphism such that the kernel is a perfect group.  Let $\mathcal{C}$ be the category whose objects are perfect homomorphisms, and whose maps are commutative squares. 
The theorem is:

There exists a class $\mathcal{F}$ of morphisms in $\mathcal{C}$ such that the category of fractions $\mathcal{C}[\mathcal{F}^{-1}]$ is equivalent to the homotopy category of pointed connected CW complexes.  

The functor from $\mathcal{C}$ to spaces is defined by the Quillen plus construction to $BG$ with respect to $P=\mathrm{Ker}(G\to H)$, and the class $\mathcal{F}$ is just the maps that become equivalences after applying the plus construction; the class $\mathcal{F}$ can also be characterized as the maps $\phi\colon (G'\to H')\to (G\to H)$ such that $H'\to H$ is an isomorphism and the induced maps $H_*(G',\phi^*M)\to H_*(G,M)$ in group homology are isos for every $H$-module $M$.
