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 $8$.

Now we consider the case that $G$ has an almost square root $A$ with $x$ as above of two-power order greater than $1$ and with $x \in Z(G)$. We claim that $x$ does have order $2$.

We consider the eigenvalues of $A^{+}$ in the regular representation of $G$. 
The trivial representation of $G$ yields the eigenvalue $+\sqrt{|G|+1}$ with multiplicity $1$. Since $G^{+}$ acts as zero and $x$ acts as a scalar on every other irreducible representation, we conclude that if $x$ has order $2^{n} >1$, then every $2^{n}$-th root of unity other than $1$ occurs with mutiplicity 
$\frac{|G|}{2^{n}}$ as an eigenvalue of $G^{+}+x$ in the regular representation
of $G$.
 Since $A^{+}$ is represented by an integer matrix in the regular representation of $G$, we conclude that algebraically conjugate eigenvalues occur with equal multiplicity, and that eigenvalues are closed under algebraic conjugation.
  Now $A^{+}$ must have some primitive $2^{n+1}$-th root of unity amongst its eigenvalues, and hence must have all $2^{n}$ primitive $2^{n+1}$-th roots of unity among its eigenvalues, with equal multiplicity $m$, say.

Consideration of the eigenvalues of $(A^{+})^{2}$ show that 
$m2^{n} = 2^{n-1}\frac{|G|}{2^{n}}.$ Thus $m = \frac{|G|}{2}.$
Hence the total multiplicity of primitive $2^{n+1}$-th roots of unity as eigenvalues of $A^{+}$ is $\frac{|G|}{2}.$ 

But if $2^{n} > 2$, a similar argument shows that the total multiplicity of primitive $2^{n}$-th  roots of unity as eigenvalues of $A^{+}$ is also $\frac{|G|}{2}$, a contradiction, since $+\sqrt{|G|+1}$ is also an eigenvalue of $A^{+}.$ Thus when $x \neq 1$ has two-power order, $x$ must indeed have order $2$.

We may now also conclude that if a general finite group $G$ of even order has an almost square root $A$ with $(A^{+})^{2} = G^{+} + x$ where $x$ is a non-identity element of $2$-power order of $G$, then $x$ has order $2$ ( using the fact that $H = C_{G}(x)$ has a similar almost square root, and $x$ is now central).

Partial answer towards problem $5$: If $G$ is a finite group with an almost square root $A$, then a Sylow $2$-subgroup of $Z(G)$ has exponent dividing $4$. 

For if $(A^{+})^{2} = G^{+} + x$, where (wlog) the order of $x$ is a power of $2$, then for every $2$-element $z \in Z(S)$, we know that 
$(zA^{+})^{2} = G^{+} + z^{2}x.$ Then by the above arguments, both $x$ and $z^{2}x$ have order dividing $2$. Hence $z^{4} = z^{4}x^{2} = 1$ and $z$ has order dividing $4$.