Let $G,H$ be torsion-free abelian groups such that $k[G] \cong k[H]$ for any field $k$. Then is it true that $G \cong H$ ?
If this is not true, then what if I change the hypothesis to $R[G]\cong R[H]$ for any non-zero commutative unital ring; is the conclusion true then ?
The group of units of $k[G]$ is exactly $k^\times\times G$. In particular, if $k=\mathbb{F}_2$, then the group of units is $G$, so $G$ can be recovered from the ring $\mathbb{F}_2[G]$.
To prove this, note that any torsion-free abelian group $G$ admits a total ordering compatible with the group structure (choose a totally ordered basis for $G\otimes\mathbb{Q}$ and use the lexicographic order). Considering $k[G]$ to be $G$-graded, suppose $u\in k[G]$ is a unit with inverse $v$. Let $a$ be the minimum degree of a homogeneous part of $u$ and $b$ be the maximum degree, and let $c$ be the minimum degree of a homogeneous part of $v$ and $d$ be the maximum degree. Then the product $uv$ has homogeneous parts of degree $a+c$ and also $b+d$. Since $uv=1$, we must have $a+c=b+d$. Since $a\leq b$ and $c\leq d$, this is only possible if $a=b$ and $c=d$, which means $u$ and $v$ are homogeneous. That is, $u$ is a unit of $k$ times an element of $G$.