A Shelah group in ZFC? In his famous paper "On a problem of Kurosh, Jonsson groups, and applications" of 1980, Shelah constructed a CH-example of an uncountable group $G$ equal to $A^{6640}$ for any uncountable subset $A\subset G$.
Let us call a group $G$
$\bullet$ $n$-Shelah if $G=A^n$ for each subset $A\subset G$ of cardinality $|A|=|G|$;
$\bullet$ Shelah if $G$ is $n$-Shelah for some $n\in\mathbb N$;
$\bullet$ almost Shelah if for each subset $A\subset G$ of cardinality $|A|=|G|$ there exists $n\in\mathbb N$ such that $A^n=G$;
$\bullet$ Jonsson if each subsemigroup $A\subset G$ of cardinality $|A|=|G|$ coincides with $G$.
$\bullet$ Kurosh if each subgroup $A\subset G$ of cardinality $|A|=|G|$ coincides with $G$.
It is clear that for any group $G$ the following implications hold:

finite $\Leftrightarrow$ 1-Shelah $\Rightarrow$ $n$-Shelah $\Rightarrow$ Shelah $\Rightarrow$ almost Shelah $\Rightarrow$ Jonsson $\Rightarrow$ Kurosh.

In the mentioned paper, Shelah constructed a ZFC-example of an uncountable Jonsson group and also a CH-example of an uncountable 6641-Shelah group.

Problem 1. Can an infinite (almost) Shelah group be constructed in ZFC?
Problem 2. Find the largest possible $n$ (which will be smaller than 6640) such that each $n$-Shelah group is finite.

This result of Protasov implies

Theorem (Protasov). Each countable Shelah group is finite.

It is easy to show that each 2-Shelah group is finite.

Problem 3. Is each 3-Shelah group finite?

 A: This is a partial answer to Problem 2 about the largest number $n$ such that every $n$-Shelah group is finite. The main result of this paper implies that this number $n<36$, at least under the Continuum Hypothesis. Combined with the inequality $n\ge 3$ proved by @YCor in his answer, we have the (sufficiently narrow) lower and upper bounds $3\le n<36$ for the number $n$.
It seems that $36$ is the smallest number $n$ that can be achieved in the Shelah's construction of an $n$-Shelah group. So, $6640$ in the original Shelah's construction of a $6640$-Shelah group can be lowered to $36$ but not less.
A: I'm not sure what makes an answer to a question with several problems of very variable difficulty a good answer :)
Anyway: 

there's no "3-Shelah" group. That is, every infinite group admits a subset $W$ such that $W^3\neq G$ and $|W|=G$. (Actually one can arrange $W\cup W^2\cup W^3\neq G$.)

Let $G$ be an infinite group. Let $A$, by Zorn, be a maximal subset such that $1\notin A\cup A^2\cup A^3$. Denote by $\langle A\rangle$ the subgroup generated by $A$, and $G^{(6)}$ the subgroup of $G$ generated by $\{g^6:g\in G\}$; clearly $G^{(6)}$ is normal in $G$.
For every $g\in G\smallsetminus A$, the maximality implies that $$1\in (A\cup\{g\})\cup(A\cup\{g\})^2\cup (A\cup\{g\})^3.$$ Since $1\notin A\cup A^2\cup A^3$, this means that 
$$1\in \{g\}\cup Ag\cup gA\cup\{g^2\}\cup  A^2g\cup AgA\cup gA^2\cup g^2A\cup gAg\cup Ag^2\cup\{g^3\}.$$
Hence one of the following holds: $g=1$ or $g^2=1$ $g^3=1$ or $g\in A^{-1}$, or $g^2\in A^{-1}$ or $g\in (A^2)^{-1}$.
Hence, $g^6\in \langle A\rangle$ for all $g\in G$; equivalently, $G^{(6)}\subset \langle A\rangle$.
In $G/G^{(6)}$, every element satisfies $x^6=1$. Since groups of exponent 6 are solvable, it follows that either $G=G^{(6)}$, or $G$ has a normal subgroup of index 2 or 3. In the last two cases, we define this subgroup as $W$. Otherwise, $\langle A\rangle=G$, that is, $A$ generates $G$. In particular, since $G$ is infinite, $|A|=|G|$, so we put $A=W$.
