My take was to assume $G$ not to be cyclic, take a nontrivial element $g\in G$ and another element $h\in G\setminus\langle g\rangle$ and show that $g^{{\rm ord}(g)/2}$ and $h^{{\rm ord}(h)/2}$ are elements of order $2$.
I cannot exclude the possibility that they're the same element. Am I on the wrong track here?
Edit: Yes, assumption is that the group is abelian.
Best Answer
If you do not assume that $G$ is abelian, then this is false: the group $Q_8$ has order $8$, but only one element of order $2$ and it is not cyclic.
However, there aren't many possibilities: every finite $p$-group with a unique subgroup of order $p$ is either cyclic of generalised quaternion.
In particular then if $G$ is abelian and as above then it is also cyclic. You can prove this directly: from the fundamental theorem of abelian groups, any abelian group decomposes into a direct product of cyclic groups, so $G=C_{n_1} \times \dots \times C_{n_k}$. From Lagrange's theorem, each $n_i$ is a power of $2$. Now any cyclic group of order a power of $2$ contains a unique element of order $2$, so each $C_{n_i}$ contains an element of order $2$. Then the only possibility is that $k=1$, so $G=C_{n_1}$ is cyclic.