I don't see how to answer this is general, but the following partial result might be of interest to you.

Theorem: If $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then $G$ does not contain $(\mathbb{Z}/p)^3$ for any prime number $p$.

Proof: Suppose $P = (\mathbb{Z}/p)^3 \leq G$. Then considered as a $\mathbb{Z}[P]$-module we still have $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[P]^{2\ell}$ for some $\ell$. We consider the Serre spectral sequence for the fibration sequence $$S \to S/P \to BP$$ in $\mathbb{F}_p$-cohomology, which has the form $E_2^{p,q} = H^p(P ; H^q(S;\mathbb{F}_p)) \Rightarrow H^{p+q}(S/G;\mathbb{F}_p)$.

The surface $S$ has genus $2+2\ell\vert P\vert$, and by Euler characteristic the surface $S/P$ therefore has genus $2+2\ell$. The spectral sequence gives an exact sequence $$0 \to H^1(P;\mathbb{F}_p) \to H^1(S/P;\mathbb{F}_p) \to H^0(P; H^1(S;\mathbb{F}_p)) \overset{d_2}\to H^2(P;\mathbb{F}_p) \to X \to 0$$ and $X$ injects into $H^2(S/G;\mathbb{F}_p)$. This exact sequence is $$0 \to \mathbb{F}_p^3 \to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{2+2\ell} \overset{d_2}\to \mathbb{F}_p^{6} \to X \to 0$$ so $X \cong \mathbb{F}_p^3$, but this cannot inject into $H^2(S/G;\mathbb{F}_p) = \mathbb{F}_p$. QED

The point of this argument is that if $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then (i) the same is true of any subgroup, and (ii) it follows from the spectral sequence argument that there are classes $a,b,c,d \in H^2(G;\mathbb{Z})$ such that $$0 \to H^i(G;\mathbb{Z}) \overset{(a,b)}\to H^{i+2}(G;\mathbb{Z}) \oplus H^{i+2}(G;\mathbb{Z}) \overset{c+d}\to H^{i+4}(G;\mathbb{Z}) \to 0$$ is exact (in fact, with any ring coefficients not just $\mathbb{Z}$). This ought to give many non-examples.

I don't see how to answer this is general, but the following partial result might be of interest to you.

Theorem: If $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then $G$ does not contain $(\mathbb{Z}/p)^3$ for any prime number $p$.

Proof: Suppose $P = (\mathbb{Z}/p)^3 \leq G$. Then considered as a $\mathbb{Z}[P]$-module we still have $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[P]^{2\ell}$ for some $\ell$. We consider the Serre spectral sequence for the fibration sequence $$S \to S/P \to BP$$ in $\mathbb{F}_p$-cohomology, which has the form $E_2^{p,q} = H^p(P ; H^q(S;\mathbb{F}_p)) \Rightarrow H^{p+q}(S/G;\mathbb{F}_p)$.

The surface $S$ has genus $2+2\ell\vert P\vert$, and by Euler characteristic the surface $S/P$ therefore has genus $2+2\ell$. The spectral sequence gives an exact sequence $$0 \to H^1(P;\mathbb{F}_p) \to H^1(S/P;\mathbb{F}_p) \to H^0(P; H^1(S;\mathbb{F}_p)) \overset{d_2}\to H^2(P;\mathbb{F}_p) \to X \to 0$$ and $X$ injects into $H^2(S/G;\mathbb{F}_p)$. This exact sequence is $$0 \to \mathbb{F}_p^3 \to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{2+2\ell} \overset{d_2}\to \mathbb{F}_p^{6} \to X \to 0$$ so $X \cong \mathbb{F}_p^3$, but this cannot inject into $H^2(S/G;\mathbb{F}_p) = \mathbb{F}_p$. QED

The point of this argument is that if $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then (i) the same is true of any subgroup, and (ii) it follows from the spectral sequence argument that there are classes $a,b,c,d \in H^2(G;\mathbb{Z})$ such that $$0 \to H^i(G;\mathbb{Z}) \overset{(a,b)}\to H^{i+2}(G;\mathbb{Z}) \oplus H^{i+2}(G;\mathbb{Z}) \overset{c+d}\to H^{i+4}(G;\mathbb{Z}) \to 0$$ is exact (in fact, with any ring coefficients not just $\mathbb{Z}$). This ought to give many non-examples.

EDIT: In fact, the argument obstructs being $\text{projective} \oplus \mathbb{Z}^2$, not just $\text{free} \oplus \mathbb{Z}^2$.

I don't see how to answer this is general, but the following partial result might be of interest to you.

Theorem: If $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then $G$ does not contain $(\mathbb{Z}/p)^3$ for any prime number $p$.

Proof: Suppose $P = (\mathbb{Z}/p)^3 \leq G$. Then considered as a $\mathbb{Z}[P]$-module we still have $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[P]^{2\ell}$ for some $\ell$. We consider the Serre spectral sequence for the fibration sequence $$S \to S/P \to BP$$ in $\mathbb{F}_p$-cohomology, which has the form $E_2^{p,q} = H^p(P ; H^q(S;\mathbb{F}_p)) \Rightarrow H^{p+q}(S/G;\mathbb{F}_p)$.

The surface $S$ has genus $2+2\ell\vert P\vert$, and by Euler characteristic the surface $S/P$ therefore has genus $2+2\ell$. The spectral sequence gives an exact sequence $$0 \to H^1(P;\mathbb{F}_p) \to H^1(S/P;\mathbb{F}_p) \to H^0(P; H^1(S;\mathbb{F}_p)) \overset{d_2}\to H^2(P;\mathbb{F}_p) \to X \to 0$$ and $X$ injects into $H^2(S/G;\mathbb{F}_p)$. This exact sequence is $$0 \to \mathbb{F}_p^3 \to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{2+2\ell} \overset{d_2}\to \mathbb{F}_p^{6} \to X \to 0$$ so $X \cong \mathbb{F}_p^3$, but this cannot inject into $H^2(S/G;\mathbb{F}_p) = \mathbb{F}_p$.

I don't see how to answer this is general, but the following partial result might be of interest to you.

Theorem: If $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then $G$ does not contain $(\mathbb{Z}/p)^3$ for any prime number $p$.

Proof: Suppose $P = (\mathbb{Z}/p)^3 \leq G$. Then considered as a $\mathbb{Z}[P]$-module we still have $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[P]^{2\ell}$ for some $\ell$. We consider the Serre spectral sequence for the fibration sequence $$S \to S/P \to BP$$ in $\mathbb{F}_p$-cohomology, which has the form $E_2^{p,q} = H^p(P ; H^q(S;\mathbb{F}_p)) \Rightarrow H^{p+q}(S/G;\mathbb{F}_p)$.

The surface $S$ has genus $2+2\ell\vert P\vert$, and by Euler characteristic the surface $S/P$ therefore has genus $2+2\ell$. The spectral sequence gives an exact sequence $$0 \to H^1(P;\mathbb{F}_p) \to H^1(S/P;\mathbb{F}_p) \to H^0(P; H^1(S;\mathbb{F}_p)) \overset{d_2}\to H^2(P;\mathbb{F}_p) \to X \to 0$$ and $X$ injects into $H^2(S/G;\mathbb{F}_p)$. This exact sequence is $$0 \to \mathbb{F}_p^3 \to \mathbb{F}_p^{2+2\ell} \to \mathbb{F}_p^{2+2\ell} \overset{d_2}\to \mathbb{F}_p^{6} \to X \to 0$$ so $X \cong \mathbb{F}_p^3$, but this cannot inject into $H^2(S/G;\mathbb{F}_p) = \mathbb{F}_p$. QED

The point of this argument is that if $H^1(S;\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ then (i) the same is true of any subgroup, and (ii) it follows from the spectral sequence argument that there are classes $a,b,c,d \in H^2(G;\mathbb{Z})$ such that $$0 \to H^i(G;\mathbb{Z}) \overset{(a,b)}\to H^{i+2}(G;\mathbb{Z}) \oplus H^{i+2}(G;\mathbb{Z}) \overset{c+d}\to H^{i+4}(G;\mathbb{Z}) \to 0$$ is exact (in fact, with any ring coefficients not just $\mathbb{Z}$). This ought to give many non-examples.