Some preliminary comments. I may add more later: Let $G$ be any finite group. If $S$ is a subset of $G$, let $S^{+} = \sum_{s \in S} s$ in the group algebra $\mathbb{C}G$. Suppose that $G$ has an almost square root $A$. If $1 \in A$, then $1 a = a 1 $ for every $a \in A$, so that $|A| = 1 $ and $|G| = 3 $, so suppose from now on that $1 \not \in A$. We have $(A^{+})^{2} = x + G^{+}$ for some $x \in G.$ Notice that $A^{+}$ commutes with $G^{+}$, so that $A^{+}$ commutes with $x$. Hence $xAx^{-1} = A$, so that the set $A$ is invariant under conjugation by $x$. Also we have $(x^{n}A^{+})^{2} = x^{2n+1} + G^{+}$ for every integer $n$, so that $x^{n}A$ is another almost square root for $G$. If the order of $x$ is not a power of $2$, then there is an integer $n$ such that the order of $x^{2n+1}$ is a power of $2$. Hence if $G$ has an almost square root, then we may arrange matters so that the order of $x$ is a power of $2$ (allowing the possibility of order $1$). In particular, we may arrange matters so that $x = 1$ when $G$ has odd order.
Now suppose that $x$ has $2$-power order greater than one. Let $a,b \in A$ with $x = ab.$ If $a,b \not \in C_{G}(x)$ then $x = a^{x}b^{x}$ is the only other factorization of $x$ as a product of two elements of $A$. If $a,b \in C_{G}(x)$ with $a \neq b$ then $x = ab = ba$ are the only expressions of $x$ as a product of two elements of $A$. If $a = b \in C_{G}(x)$ then there is exactly one more expression $c^{2} = x$ with $c \in A$ (since $x$ can't then be a product of two elements of $A$, each outside $C_{G}(x)).$
Now let $H = C_{G}(x)$ and $D = A \cap H.$ Suppose that $ d \in H $ with $d \neq x.$ Then $d = uv$ for some $u,v \in A$ and $u,v$ are unique. Then $d = d^{x} = u^{x}v^{x}$ and $u^{x},v^{x} \in A$. Hence $u = u^{x}$ and $v = v^{x}.$ Thus $u, v \in D$.
Now we have $(D^{+})^{2} = H^{+} \pm x $. If the product is $H^{+} + x,$ then $D$ is an almost square root for $H = C_{G}(x).$
Note also that $1 = uv$ for some $u,v \in D$. If $u \neq v$ we obtain a contradiction since both $1$ and $x$ occur twice in $A.A$. Hence $u = v$, so that $u$ is an involution, since $1 \not \in A$. Furthermore, $u$ is the unique involution in $A$.