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Take any group $G$ (non-abelian if you like) that has a presentation $G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_{s} \rangle$ with the same number of generators and relations.

Form a CW-complex $X$ with one $0$-cell, $s$ 1-cells (representing the generators $g_i$) and $s$ $2$-cells that represent the relations. Then $\pi_1(X) = G$ and the cellular chain complex that computes $H_{\ast}(X)$ looks as follows: $$\mathbb Z^{s} \xrightarrow{\partial} \mathbb Z^s \xrightarrow{0} \mathbb Z$$ The differential $\partial$ can of course be calculated easily by abelianizing the relations $r_i$. If $G$ is finite, then $H_1(X;\mathbb Q) = G^{\text{ab}} \otimes \mathbb Q = 0$, hence the induced map $\partial \otimes \mathbb Q$ is surjective, hence injective, hence also $\partial$ itself is injective, hence $H_2(X;\mathbb Z) = 0$. Thus, $X$ is as desired.

It remains to undertstand which (finite) groups (if any?) admit such a presentation.

Take any group $G$ (non-abelian if you like) that has a presentation $G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_{s} \rangle$ with the same number of generators and relations.

Form a CW-complex $X$ with one $0$-cell, $s$ 1-cells (representing the generators $g_i$) and $s$ $2$-cells that represent the relations. Then $\pi_1(X) = G$ and the cellular chain complex that computes $H_{\ast}(X)$ looks as follows: $$\mathbb Z^{s} \xrightarrow{\partial} \mathbb Z^s \xrightarrow{0} \mathbb Z$$ The differential $\partial$ can of course be calculated easily by abelianizing the relations $r_i$. If $G$ is finite, then $H_1(X;\mathbb Q) = G^{\text{ab}} \otimes \mathbb Q = 0$, hence the induced map $\partial \otimes \mathbb Q$ is surjective, hence injective, hence also $\partial$ itself is injective, hence $H_2(X;\mathbb Z) = 0$. Thus, $X$ is as desired.

It remains to undertstand which finite groups admit such a presentation. In line with mme's great comment above, $2I$ is such an example.

Take any group $G$ (non-abelian if you like) that has a presentation $G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_s \rangle$ with the same number of generators and relations.

Form a CW-complex $X$ with one $0$-cell, $s$ 1-cells (representing the generators $g_i$) and $s$ $2$-cells that represent the relations. Then $\pi_(X) = G$ and the cellular chain complex that computes $H_{\ast}(X)$ looks as follows: $$\mathbb Z^s \xrightarrow{\partial} \mathbb Z^s \to \mathbb Z$$ The differential $\partial$ can of course be calculated easily by abelianizing the relations $r_i$. If $G$ is finite, then $H_1(X;\mathbb Q) = G^{\text{ab}} \otimes \mathbb Q = 0$, hence $\partial \otimes \mathbb Q$ is surjective and hence injective, hence also $\partial$ itself is injective, hence $H_2(X;\mathbb Z) = 0$. Thus, $X$ is as desired.

It remains to undertstand which (finite) groups $G$ admit such a presentation.

Take any group $G$ (non-abelian if you like) that has a presentation $G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_{s} \rangle$ with the same number of generators and relations.

Form a CW-complex $X$ with one $0$-cell, $s$ 1-cells (representing the generators $g_i$) and $s$ $2$-cells that represent the relations. Then $\pi_1(X) = G$ and the cellular chain complex that computes $H_{\ast}(X)$ looks as follows: $$\mathbb Z^{s} \xrightarrow{\partial} \mathbb Z^s \xrightarrow{0} \mathbb Z$$ The differential $\partial$ can of course be calculated easily by abelianizing the relations $r_i$. If $G$ is finite, then $H_1(X;\mathbb Q) = G^{\text{ab}} \otimes \mathbb Q = 0$, hence the induced map $\partial \otimes \mathbb Q$ is surjective, hence injective, hence also $\partial$ itself is injective, hence $H_2(X;\mathbb Z) = 0$. Thus, $X$ is as desired.

It remains to undertstand which (finite) groups (if any?) admit such a presentation.

Isn't the projective plane $\mathbb RP^2$ such an example? It has $\pi_1 = \mathbb Z/2$ and $H_i = 0$ for all $i> 0$.

Take any group $G$ (non-abelian if you like) that has a presentation $G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_s \rangle$ with the same number of generators and relations.

Form a CW-complex $X$ with one $0$-cell, $s$ 1-cells (representing the generators $g_i$) and $s$ $2$-cells that represent the relations. Then $\pi_(X) = G$ and the cellular chain complex that computes $H_{\ast}(X)$ looks as follows: $$\mathbb Z^s \xrightarrow{\partial} \mathbb Z^s \to \mathbb Z$$ The differential $\partial$ can of course be calculated easily by abelianizing the relations $r_i$. If $G$ is finite, then $H_1(X;\mathbb Q) = G^{\text{ab}} \otimes \mathbb Q = 0$, hence $\partial \otimes \mathbb Q$ is surjective and hence injective, hence also $\partial$ itself is injective, hence $H_2(X;\mathbb Z) = 0$. Thus, $X$ is as desired.

It remains to undertstand which (finite) groups $G$ admit such a presentation.