Let $a=2x+3y$ , and $b=x-y$ . Then, if $g=na+mb=sx+yt$ , then $g=(2n+m)x+(3n-m)y$ , hence, $2n+m=s$ and $3n-m=t$ , which has no more than 1 solution, hence, $\{a,b\}$ forms a basis for the free subgroup generated by them.

However, I am not sure the subgroup is actually the whole group $G$ . For example, even if we want to express $x$ as a linear combination of $a$ and $b$ , we end up with the system of equations $2n+m=1$ and $3n-m=0$ , which implies $n=1/5$ , but both $n$ and $m$ have to be integers.

For example, consider the set of all integer points of the $x,y$ -plane with usual addition. It is a free abelian group having a basis $x=(1,0)$ and $y=(0,1)$ . Then, the points $(2,3)$ and $(1,-1)$ generate all points of the form $(2n+m,3n-m)$ , in particular, the sum of the coordinates of these points is always divisible by $5$ . So, not every point can be express as a sum of elements of the groups generated by $a$ and $b$ .

I believe, this is a typo, and either the second one should be $x+y$ , or the first one should be $2x-3y$ . So, if we have $\{a,b\}=\{2x+3y,x+y\}$ , then $g=na+mb=sx+ty$ implies $2n+m=s$ and $3n+m=t$ , or $n=t-s$ , and $m=3s-2t$ . Since $s$ and $t$ are determined uniquely by $g$ , we conclude that so do $n$ and $m$ . Moreover, $na=0$ implies $n=0$ , and similarly for $b$ . Hence, $G$ is a free group having a basis $\{a,b\}=\{2x+3y,x+y\}$ . Similarly, for $\{2x-3y,x-y\}$ .