One can define $\mathrm{Ext}$-groups in the category of abelian groups (not $\mathbb{Z}[G]$-modules) and group cohomology in very similar ways. The second, group cohomology, can be computed in the homotopy category of topological spaces: we can take a group $G$, turn it into a topological space $\mathrm{K}(G,1)$ by geometric realisation, and we have $$ \mathrm{H}^n(G,\mathbb{Z}) = \left [ \mathrm{K}(G,1), \mathrm{K}(\mathbb{Z},n) \right ].$$
What about the first? In this case, working in the category of chain complexes of abelian groups, one gets $\mathrm{Ext}$-groups instead (which fits in the above framework by considering the homology of a chain complex to be its homotopy groups).
I'd like to reconcile this with the principle that considering abelian groups is akin to stabilising: chain complexes of abelian groups correspond to certain kinds of infinite loop spaces, inside the whole category of topological spaces. Is there a stabilisation process that relates $\mathrm{Ext}$-groups of abelian groups and group cohomology of abelian groups that underlies the above picture?
Consider for instance the cohomology of $\mathbb{RP}^\infty$, and contrast with $\mathrm{Ext}$-groups.
For $i>0$ even: $$ \mathrm{H}^i(\mathbb{RP}^\infty,M) = M/2M, \qquad \mathrm{Ext}^1(\mathbb{Z}/2\mathbb{Z},M) = M/2M, $$ and for $i>0$ odd: $$ \mathrm{H}^i(\mathbb{RP}^\infty,M) = M[2], \qquad \mathrm{Hom}(\mathbb{Z}/2\mathbb{Z},M) = M[2]. $$
How does this work in general, for $\mathrm{K}(A,1)$ with $A$ an abelian group (or perhaps even for more general $\mathrm{K}(A,n)$)?
Edit: Removed incorrect objects obtained by the bar construction. The off-by-one in the above two equations is a manifestation of my error, maybe someone can explain how to use bar constructions attached to an adjunction to construct classifying spaces.
As you have noticed, there is an irritating degree shift that pervades this discussion. Consider the simplicial set $B G$ whose $n$-simplices are $n$-tuples of elements of $G$, with face and degeneracy operators are in the bar construction. (In fact, it is the bar construction induced by the free–forgetful adjunction for $G$-sets.) In other words, $B G$ is the nerve of $G$ considered as a one-object groupoid, and it is well-known that the nerve of a groupoid is always a Kan complex. However, it is not a contractible Kan complex in general: indeed, the geometric realisation of $B G$ is a $K (G, 1)$ space.
We can then take the décalage of $B G$ to get a simplicial set $E G$ whose $n$-simplices are the $(n + 1)$-tuples of elements of $G$. If we choose the appropriate convention, $E G$ will be the nerve of the groupoid whose objects are the elements of $G$ and whose morphisms $x \to y$ are the elements $g$ such that $g x = y$. In particular, $E G$ is a contractible Kan complex, and we have a Kan fibration $p : E G \to B G$ whose fibre is canonically identified with the underlying set of $G$. In particular this exhibits the geometric realisation of $B G$ as a classifying space for $G$.
Now, consider the associated unnormalised chain complexes. On the one hand, for any abelian group $A$, $$H^* (K (G, 1), A) \cong H^* (\mathrm{Hom}_\mathbb{Z} (\mathbb{Z} B G, A))$$ by the isomorphism of singular and cellular cohomology. On the other hand, $\mathbb{Z} E G$ is (by Dold–Kan) a resolution of $\mathbb{Z}$ that is degreewise a free right $\mathbb{Z} G$-module. Thus, $$H^* (G, A) = \mathrm{Ext}_{\mathbb{Z} G}^* (\mathbb{Z}, A) \cong H^* (\mathrm{Hom}_{\mathbb{Z} G} (\mathbb{Z} E G, A))$$ On the other hand, $\mathbb{Z} E G$ is isomorphic to $\mathbb{Z} B G \otimes_\mathbb{Z} \mathbb{Z} G$, so $\mathrm{Hom}_{\mathbb{Z} G} (\mathbb{Z} E G, A) \cong \mathrm{Hom}_{\mathbb{Z}} (\mathbb{Z} B G, A)$, which proves $$H^* (G, A) \cong H^* (K (G, 1), A)$$ as is well known.
The business with $\mathrm{Ext}_\mathbb{Z}^*$ is quite different, but there is a curious coincidence worth pointing out. In low degrees, the chain complex $\mathbb{Z} B G$ has differentials given by $(g_0, g_1) \mapsto g_0 - g_0 g_1 + g_1$ and $g \mapsto 0$; so the cochain complex $\mathrm{Hom}_\mathbb{Z} (\mathbb{Z} B G, A)$ has differentials given by $f \mapsto 0$ and $f \mapsto ((g_0, g_1) \mapsto f (g_0) - f (g_0 g_1) + f (g_1))$. Thus: $$H^1 (K (G, 1), A) \cong \mathrm{Hom} (G, A)$$ This is homotopy-theoretically unsurprising: group homomorphisms $G \to H$ can be canonically identified with pointed homotopy classes of pointed maps $K (G, 1) \to K (H, 1)$, which in turn can be identified with elements of $H^1 (K (G, 1), H)$.
Is tempting to conjecture that $\mathrm{Ext}^* (G, A) \cong H^{* + 1} (K (G, 1), A)$ in general; sadly, this claim is false. Let $\mathbb{Z} B G [1]$ be the chain complex defined by $\mathbb{Z} B G [1]_n = \mathbb{Z} B G_{n + 1}$. Since the differential $\mathbb{Z} B G_1 \to \mathbb{Z} B G_0$ is zero, we immediately deduce $$H_* (\mathbb{Z} B G [1]) \cong H_{* + 1} (\mathbb{Z} B G) \cong H_{* + 1} (K (G, 1), \mathbb{Z})$$ and therefore $\mathbb{Z} B G [1]$ is not a free resolution of $H_1 (K (G, 1), \mathbb{Z}) \cong G / [G, G]$ in general! Unsurprisingly, we can find examples where $\mathrm{Ext}_\mathbb{Z}^* (G, A)$ and $H^{* + 1} (K (G, 1), A)$ are genuinely different. For instance, given $n > 0$, $$\mathrm{Ext}_\mathbb{Z}^n (\mathbb{Z}^{n+1}, \mathbb{Q}) = 0$$ because $\mathbb{Q}$ is an injective $\mathbb{Z}$-module, whereas $$H^{n+1} (K (\mathbb{Z}^{n+1}, 1), \mathbb{Q}) \cong H^{n+1} (T^{n+1}, \mathbb{Q}) \cong \mathbb{Q}$$ because $K (\mathbb{Z}^{n+1}, 1) \simeq K (\mathbb{Z}, 1)^{n+1}$, $K (\mathbb{Z}, 1)$ is homotopy equivalent to the circle $S^1$, and the torus $T^{n+1}$ is a connected compact orientable manifold of dimension $n + 1$.