Your approach is not wrong at all , but it needs some clarity. And also a better understanding of what have u constructed. So,
As u have shown, $\Bbb{H}$= $<x>$ , where o(x)=4. And as $\Bbb{H}$ , is cyclic , order of it is also 4 . And [ $\Bbb{S}$ : $\Bbb{H}$ ]=2, thus $H\trianglelefteq S$ . Now as identity is unique and it is in $\Bbb{H}$ , then no element from $\Bbb{S}$ \
$\Bbb{H}$ will have order 1 .
Now when we are considering g, it can be picked up either from $\Bbb{H}$ or from ,
$\Bbb{S}$ \ $\Bbb{H}$.
Now if it's taken from $\Bbb{H}$ , as $\Bbb{H}$ is cyclic, $\Bbb{H}$ must be commutative, thus ,
$\forall$ g in $\Bbb{H}$ ,
g $x^{2}$ = $x^{2}$g holds.
Now when g is in $\Bbb{S}$ \ $\Bbb{H}$, o(g)≠1 . If it is 4 then the same argument comes . Now if o(g)=2=o( $x^{2}$ ), then they will commute and ,
g $x^{2}$= $x^{2}$ g holds .
And at last if, o(g)=8 , Then as o(g)=o( $\Bbb{S}$ ) , $\Bbb{S}$ will be cyclic. Thus it will be commutative. And then also,
g $x^{2}$ = $x^{2}$g will hold.
So , $\forall$ g in $\Bbb{S}$ , the relation will hold .
Well, to understand this ongoing argument more clearly, let's have an example. Let's consider the dihedral group $D_{4}$ which has 8 elements and they are , {r, $r^{2}$ , $r^{3}$ , $r^{3}$ , b, b', d, d '} . Here,
b and b ' represent reflection about diagonals .
d and d ' represent reflection about the lines joining middle points of opposite arms of the square .
r= $R_{90}$ ,
$r^{2}$= $R_{180}$ ,
$r^{3}$= $R_{270}$ ,
$r^{4}$= $R_{360}$ .
Now as, b=r $\ast$ b $\ast$ $r^{-1}$
And, d= r $\ast$ d' $\ast$ $r^{-1}$ .
We can construct 2 subgroups of $D_{4}$ , which are ,
$\Bbb{T}$ = { e, b, b', $r^{2}$ }
$\Bbb{F}$ = { e, d, d' , $r^{2}$ }
Now , $\Bbb{T}$ $\trianglelefteq$ $D_{4}$
And , $\Bbb{F}$ $\trianglelefteq$ $D_{4}$
And most astonishingly these two are cyclic groups also . Additive cyclic groups , having the generator , $r^{2}$.
Now there leaves only two proper subgroups of $D_{4}$ , one is the group contained first four elements i.e {r, $r^{2}$ , $r^{3}$ , $r^{3}$ } , Which we have discussed earlier. So there left the group only {e, $r^{2}$ } .
And from all these one thing that we are sure about is , $r^{2}$ is only the element ( of order 2) which commutes with every element of $D_{4}$
So, $r^{2}$ $\in$ $\Bbb{Z}$( $D_{4}$ )
And it ends here .......( Well, it's just the beginning ! )....