Definition. A finite group $G$ is called squared (resp. almost squared) if there exists a subset $A\subseteq G$ such that $G=\{ab:a,b\in A\}$ and $|G|=|A|^2$ (resp. $|G|=|A|^2-1$). Such a set $A$ will be called an (almost) square root of $G$.
According to the answers to this MO-post, no nontrivial finite group is squared.
In contrast, nontrivial almost squared finite groups do exist. The simplest one is the 3-element cyclic group $C_3$. Any 2-element subset of $C_3$ is an almost square root of $C_3$.
A less trivial example is the dihedral group $D_8=\langle a,b\;|\;a^4=b^2=1,\;bab=a^3\rangle$ with almost square root $A=\{a,b,ba\}$.
Three other examples of almost squared groups (found by GAP) are:
$\bullet$ the symmetric group $S_4$;
$\bullet$ the general linear group $GL(2,3)$ of non-degenerate $2\times 2$ matrices over the 3-element field,
$\bullet$ the symmetric group $S_5$.
Problem 1. Find more examples of almost squared finite group. Are there infinitely many almost squared finite groups?
Remark. Using GAP, Voldymyr Gavrylkiv established that among groups of order $<168$ the only almost squared groups are the groups $C_3$, $D_8$, $S_4$, $GL(2,3)$, and $S_5$. Those groups have orders 3, 8, 24, 48, and 120, respectively. It is interesting that no almost squared group of order 80 exists.
Problem 2. What can be said about the structure of almost squared groups?
Remark. Alex Ravsky observed that for an almost square root $A$ of an almost squared group $G$, the center $Z(G)$ of $G$ is almost contained in the set $A^2=\{a^2:a\in A\}$ in the sense that $Z(G)\setminus A^2$ contains at most one element. So, $|Z(G)|\le|A|+1=1+\sqrt{|G|+1}$, which implies that the unique almost squared commutative group is $C_3$.
The only known (at the moment) almost squared noncommutative groups $D_8$, $S_4$ and $GL(2,3)$ have even cardinality.
Problem 3. Is the cardinality of any almost squared noncommutative group even?
Remark. It can be shown that all noncommutative groups of odd order $<675=3^3\times 5^2$ are not almost squared.
Problem 3'. Is there an almost squared group among groups of order $675$?
Problem 4. Let $A$ be an almost square root of an almost squared non-commutative finite group $G$. Are there distinct elements $a,b\in A$ such that $a^2=b^2=g$ for some $g\in G$ such that:
$\bullet$ $g\in Z(G)$?
$\bullet$ $g^2\in Z(G)$?
$\bullet$ $g^2=1$?
$\bullet$ $g=1$?
$\bullet$ $g$ has order $\le 3$?
Remark. For any almost square root in any of three known almost squared noncommutative groups $D_8,S_4,GL(2,3)$, there exist distinct elements $a,b\in A$ such that $a^2=b^2=1$. In these group the center contains at most two elements.
The following problem was suggested by @LSpice in his comment.
Problem 5. Let $G$ be an almost squared noncommutative group. Is $z^2=1$ for any central element of $G$?
Best Answer
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 $H = C_{G}(x).$
Suppose then $(D^{+})^{2} = H^{+}-x.$ Note that $H$ has even order as $x \in H$. In the regular representation of $H$, the element $H^{+}- x$ has the eigenvalue $|H| - 1$ with multiplicity $1$, and $D^{+}$ has the eigenvector $|D|$ on that eiegenspace. Hence $|H|-1$ is the square of an odd integer. Thus $|H| \equiv 2$ (mod $8$). In that case, $\langle x \rangle$ is a Sylow $2$ subgroup of $H = C_{G}(x) = N_{G}(\langle x \rangle)$, so is also a Sylow $2$-subgroup of $G$. This is a standard group-theoretic argument, but I explain it for non-group theorists : Note that $x$ lies in some Sylow $2$-subgroup $S$ of $G$. Suppose that $|S| > 2.$ If $x \in Z(S)$, then $S \leq C_{G}(x),$ so that $|S|$ divides $|H|$ which is not the case. If $x \not \in Z(S)$, then $\langle x, Z(S) \rangle \leq H$ and $|\langle x, Z(S) \rangle | \geq 2|\langle x \rangle |,$ a contradiction.
However, we can now reach a contradiction. If $G$ is a group of even order such that $A$ is an almost square root for $G$, then $|A|^{2} = |G| + 1$, so that $|G|+1$ is the square of an odd integer. Hence $|G| + 1 \equiv 1$ (mod $8$), so that $|G|$ is divisible by $8$.
But in the present case, $|G|$ is not divisble by $4$. Hence the situation $(D^{+})^{2} = H^{+} - x$ can not occur when $x$ is a non-identity element of $2$-power order of $G$, and $G$ has an almost square root $A$ such that $(A^{+})^{2} = G^{+} + x$, where $H = C_{G}(x).$
In particular, we may deduce that if $G$ is a finite group of even order with an almost square root $A$ such that $(A^{+})^{2} = G^{+} + x$ for some non-identity element $x$ whose order is a power of $2$, then $D = A \cap C_{G}(x)$ is an almost square root for $H = C_{G}(x).$
To recap : if $G$ is a finite group with an almost square root $A$, and we have $(A^{+})^{2} = G^{+} + x$ for some non-identity element $x$, then we can find an integer $n$ such that $x^{n}A$ is an almost square root for $G$ with $(x^{n}A^{+})^{2} = G^{+} + x^{2n+1}$, where the order of $x^{2n+1}$ is a power of $2$ (possibly $1$). This reduces to considering the case that the order of $x$ is a power of $2$ (possibly $1$). If the order of $x$ is a power of $2$ greater than one, the group $C_{G}(x)$ has the almost square root $D = A \cap C_{G}(x).$
Hence a key case to understand is when $x$ is a central element of $G$ whose order is a power of $2$ (possibly $1$).
Continued : Notice also that if $x$ has even order, then $|C_{G}(x)|$ is a group of even order with an almost square root, so that $|C_{G}(x)|$ has order divisible by $8$.