Not every set of elements of finite order is a subgroup. What is true though is that the product and inverse of elements with finite order also has finite order, so the collection of ALL elements of finite order is a subgroup.
For product of elements of finite order take let $g$ and $h$ be of order $n$ and $m$ respectively. Try to find some large number $k$ such that $(gh)^k=e$.
Hint: in an abelian group exponents distribute over multiplication.
I'll leave proving that inverses of elements of finite order have finite order to you.
Expanding on my comment, let us apply Burnside's transfer theorem to give a nice and conceptual proof of this (and some more general facts).
Theorem (Burnside): Let $G$ be a group with a $p$-Sylow subgroup $P$ satisfying $P\leq Z(N_G(P))$. Then $G$ has a normal subgroup $N$ of index $|P|$ (i.e. a normal $p$-complement).
Theorem (normalizer/centralizer): Let $H$ be a subgroup of a group $G$. Then $C_G(H)$ is a normal subgroup of $N_G(H)$ and $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$.
I will lave the proofs as exercises. The second one is a good exercise while the first one is a terrible exercise as it is way too hard (the two proofs I know of either use a lot of stuff about transfer or uses a lot of character theory).
Anyway, using these, we can now show that if $p$ is the smallest prime dividing the order of the group (and in the case at hand, we are considering $p=2$ which will certainly be smallest), and if the $p$-Sylow subgroup is cyclic (which is precisely the assumption), then $G$ has a normal $p$-complement, meaning that the elements of order not divisible by $p$ form a normal subgroup.
To apply Burnside's result here, we need to show that $P\leq Z(N_G(P))$, but this is the same as saying that $N_G(P) = C_G(P)$, and we can show that by showing that $N_G(P)/C_G(P)$ is trivial.
Now, by the normalizer/centralizer theorem, this quotient is isomorphic to a subgroup of $\operatorname{Aut}(P)$, which we know has order $\varphi(|P|)$ since $P$ is cyclic. But this number is coprime to $\frac{|G|}{|P|}$, whereas the order of $N_G(P)/C_G(P)$ must divide $\frac{|G|}{|P|}$ since $C_G(P)$ contains $P$. This precisely implies that $N_G(P) = C_G(P)$ as we wanted, which finishes the proof.
Best Answer
Say we have an abelian group with two elements $a, b$ of order $o_a, o_b$ respectively, and let $d = \operatorname{lcm}(o_a, o_b)$. Then $d\cdot(a+b) = 0$, so the order of $a+b$ must divide $d$. If $o_a$ and $o_b$ are both odd, then $d$ is odd, and thus the order of $a + b$ must also be odd.
There is nothing special about "odd" here, we can use "coprime to $n$" for any natural number $n$ in place of "odd".