This won't work. I want to show that we can't take square roots in $A(\mathbb R)$. My function will be of the type
$$
f(x) = \sum h_n \varphi\left( \frac{x-a_n}{L_n}\right) ,
$$
and here the individual summands will have disjoint supports. I will take $h_n\in\ell^2$, $h_n\notin\ell^1$. Since the $L^1$ norm of the Fourier transform of $\psi((x-a)/L)$ is independent of $a,L$, the first property will make sure that $\widehat{f^2}\in L^1$, that is $f^2\in A(\mathbb R)$.

So it is now enough to find $\varphi\ge 0$ and $h_n,a_n,L_n$ such that $f\notin A(\mathbb R)$. Fix a $\varphi\ge 0$ that is supported by $[0,1]$, with $\varphi'(0)>0$ and $\varphi$ is smooth otherwise. Then (after normalizing suitably) we will have that $|\widehat{\varphi}(t)|=1/t^2+O(t^{-3})$, and of course $\widehat{\varphi}$ is bounded. Thus the Fourier transform of the $n$ summand of (1) is of the order
$$
\frac{h_n}{L_n x^2} + O(h_nL_n^{-2}x^{-3}) ,
$$
and it is bounded by $Ch_nL_n$. Thus for sufficiently large $B$, this summand on its own on the interval $B/L_n\le x\le 2B/L_n$ would make a contribution $\gtrsim h_n/B$ to the $L^1$ norm of $\widehat{f}$.

We will be done if we can make sure that the other summands cannot completely cancel out this contribution. The summands with $k>n$ cannot contribute more than $CBh_{n+1}L_{n+1}/L_n$, and the ones with $k<n$ can be treated as above, with $h_n/L_n$ replaced by $h_k/L_k$. So all competing contributions are much smaller if we take $L_n$'s that converge very rapidly to zero.