Convoluted definition of a set: $H(S) = \{x\in G\mid\exists n\in\Bbb N,\exists\{x_1, x_2, ..., x_n\}\subseteq S\cup S^{-1}, x = x_1 ...x_n\}$

$H(S)$ is the set of all elements $g$ of $G$ such that $g$ belongs to some finite subset of $S\cup S^{-1}$?

No. $H(S)$ is the set of all elements $g$ of $G$ such that $g$ can be expressed as a finite product of elements of $S\cup S^{-1}$.

Maybe lets have a look at example. Consider $G=\{0,1,2,3\}$ with addition modulo $4$ as group operation, i.e. $G=\mathbb{Z}_4$. Now let $S=\{2\}$. Since we are dealing with addition, then $S^{-1}=\{2\}$ because $2+2=0$ in our group. Therefore $S\cup S^{-1}=\{2\}$.

Next $H(S)$ is the set of all elements $g$ of $G$ such that $g=x_1+\cdots +x_n$ for some $x_1,\ldots,x_n\in S\cup S^{-1}$. Since our last set has exactly one element then only possibilities for $g$ are: $2$, $2+2$, $2+2+2$, and so on... which we can write in a compact way as $n\cdot 2$ for some $n\in\mathbb{N}$. By removing duplicates (remember that $2+2=0$) this gives us

$$H(S)=\{0,2\}$$