Can a group have two different subgroups of index $2$?

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(\mathbb{Z}/2\mathbb{Z})\times (\mathbb{Z}/4\mathbb{Z})$.

It is maybe worth noticing the following. Let $G$ be a group and let $I_2(G)=\#\{H \lt G: |G:H|=2\}$, be the number of subgroups of index $2$.

Theorem (Crawford, Wallace, 1975) Let $G$ be a group and $n$ a non-negative integer. Then $I_2(G)=n$ if and only if $n=2^k-1$ for some non-negative integer $k$.

See On the Number of Subgroups of Index Two-An Application of Goursat's Theorem for Groups, R. R. Crawford and K. D. Wallace, Mathematics Magazine Vol. 48, No. 3 (May, 1975), pp. 172-174.

From this it follows that $I_2(G) \equiv 1$ or $3$ mod $6$, and in particular, $I_2(G) \neq 2$.

Another observation (see also the quoted paper): the following are equivalent.

(a) A group $G$ has a unique subgroup of index $2$.
(b) $G$ cannot be expressed as the union of 3 different subgroups.
(c) $G$ does not have a quotient isomorphic to Klein's group $V_4$.

No way is it unique (generally). The group $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ has three subgroups of index $2$: $\mathbb{Z}_2 \times\{0\}$, the vice versa, and $\{(0,0), (1,1)\}$. You can use the same idea to cook up plenty of other examples.