$H$ is a subgroup of index $2$ in finite group $G$, then every left coset of $H$ is a right coset as well

The (left) cosets of $H$ partition the group $G$; since there are two cosets and one of them is $H$, that means that, whatever the other one is (call it $\mathcal{C}$), as a set, you have that $H\cap\mathcal{C}=\emptyset$ and $H\cup \mathcal{C}=G$. These two together tell you that the other one must be equal, as a set, to the complement of $H$ inside of $G$; that is, $\mathcal{C}=G-H$. So the two cosets are $H$ and $G-H$. Of course, $G-H$ can be written as $xH$ for some $x$; in fact, for any $x\notin H$ you have $xH = G-H$.

But the exact same argument holds for right cosets: one of them is $H$, and the other one must be, as a set, the complement of $H$. So the two right cosets are $H$ and $G-H$, and $G-H$ can be written as $Hy$ for some $y$; in fact, for any $y\notin H$ you have $Hy=G-H$.

Now note that in both cases, the cosets are: 1. $H$, the coset of the elements of $H$; and 2. $G-H$, the coset of the elements not in $H$.

So each of the two left cosets is also a right coset and vice-versa.

You say "Also since there is only one possible coset, this means that for all elements, a,b in G, then aH = bH and Ha = Hb, since each element in G acting on the subgroup must produce the same set." but this is not true. Let a be the identity and b not be in H.

But earlier you say "I know this means that there are two cosets. I also know that one must be H itself." And if all cosets are the same size, what can you say about the one that isn't H?

For a problem like this one, it helps to know the principle that

Any two left (right) cosets are either disjoint or equal.

This may be proved as follows: suppose $g_1H$ and $g_2H$ are left cosets of $H$ in $G$, and $g_1H \cap g_2 H \ne \emptyset$; then there must be some $h_1, h_2 \in H$ with

$g_1 h_1 = g_2 h_2; \tag 1$

thus

$g_1 = g_2 h_2 h_1^{-1}; \tag 2$

i.e., we have

$h = h_2 h_1^{-1} \in H \tag 3$

with

$g_1 = g_2 h; \tag 4$

now if

$g_1 h_3 \in g_1H, \tag 5$

then by (4),

$g_1 h_3 = g_2 h h_3 \in g_2 H, \tag 6$

which shows that

$g_1 H \subset g_2 H; \tag 7$

likewise, the roles of $g_1$ and $g_2$ may reversed in the above to show that

$g_2 H \subset g_1 H; \tag 8$

from (7) and (8) we conclude that

$g_1 H = g_2 H. \tag 9$

A similar demonstration works for right cosets. We note that this Principle holds for any group $G$ and subgroup $H$, whether finite or not, and whether $[G:H]$ is finite or not.

We exploit said Principle in the present question as follows: if $e \in G$ is the identity element, then clearly

$eH = He = H, \tag{10}$

and if

$g \in G \setminus H, \tag{11}$

then

$gH \ne H, \tag{12}$

since

$g = ge \in gH, \tag{13}$

but by (11),

$g \notin H; \tag{14}$

therefore,

$gH \ne H, \tag{15}$

and it thus follows from our Principle that

$gH \cap H = \emptyset. \tag{16}$

Since $[G:H] = 2$, it follows that $H = eH$ and $gH$ are the only two left cosets of $H$ in $G$; thus

$G = H \cup gH; \tag{17}$

in a similar manner we may also see that

$G = H \cup Hg, \; H \cap Hg = \emptyset, \tag{18}$

and we conclude from (16)-(18) that

$gH = Hg. \tag{19}$

We see that the above demonstration uses notions of basic set theory as well as the fact that there are only two distinct left or right cosets, which are disjoint, and one of which is always $eH = He = H$. And obviously, a good amount of logic.