This cannot work. For example, take $G=\mathbb R$ and consider a function $f(x)=x^{\alpha}$ near $x=0$ and $f\ge 0$ is smooth and compactly supported otherwise. Then $\widehat{f}\sim$ near infinity and thus the Fourier transform is not integrable, so $f\notin A(\mathbb R)$.
However, $g=f^n$ has enough smoothness to make $\widehat{g}\in L^1$. So $g\ge 0$ is in $A(\mathbb R)$, but we can not take a square root without leaving this space.