This won't work. I want to show that we can't take square roots in $A(\mathbb R)$. My counterexample will be of the type $$ f(x)=\sum h_n \varphi\left(\frac{x-a_n}{L_n}\right) , $$ where $\varphi$ has compact support and the supports of distinct terms from this sum are disjoint.
I will then take $h_n\in\ell^2$, $h_n\notin\ell^1$. Since the $L^1$ norm of the Fourier transform of $\psi(x/L)$ does not depend on $L$, the first condition makes sure that $\widehat{f^2}\in L^1$, that is, $f^2\in A(\mathbb R)$.
By taking wildly different, rapidly decreasing scales $L_n\to 0$, I can then mostly avoid interactions between the individual summands when evaluating $\widehat{f}$ and achieve that $\widehat{f}\notin L^1$. More specifically, take a $\varphi\ge 0$ that is supported by $[0,1]$, $\varphi'(0)>0$ and $\varphi$ is smooth otherwise. Then (for a suitable multiple) $|\widehat{\varphi}(t)| = t^{-2} + O(t^{-3})$ and thus the Fourier transform of the $n$th summand of (1) is of the order $$ \frac{h_n}{L_n} $$