I know this problem has been asked a couple times in the site, but this is a proof verification question.
Attempt:
Let $G$ be an Abelian group of odd order. We know that $\forall g\in G$, $\vert g\vert=n$ divides $\vert G\vert$. Thus $n$ cannot be even. Since $\forall g\in G$, $\langle g\rangle \leq G$ is a cyclic group; $\langle g\rangle=\{e,g,g^2,\cdots,g^{n-1}\}$ observe that
\begin{align}
egg^2\cdots g^{n-2}g^{n-1}=(gg^{n-1})(g^2g^{n-2})\cdots (g^{\frac{n-1}{2}}g^{\frac{n+1}{2}})=e\quad (*)
\end{align}
Note that $n$ not being even implies $\frac{n-1}{2},\frac{n+1}{2}\in\Bbb{N}$. Since $G=\bigcup_{i=1}^{\vert G\vert}\langle g_i \rangle$, the product of all elements in $G$ is equivalent to the form $e(g_{1}g_{1}^{n-1}\cdots g_{1}^{\frac{n-1}{2}}g_{1}^{\frac{n+1}{2}})\cdots (g_{z}g_{z}^{n-1}\cdots g_{z}^{\frac{n-1}{2}}g_{z}^{\frac{n+1}{2}})$ where $z=\vert G\vert$. Then as shown in $(*)$ this is equal to the identity.
Best Answer
The idea behind your proof is correct, but there are some wee peccadilloes in the mix. When you write out the long product $$ e(g_{1}g_{1}^{n-1}\cdots g_{1}^{\frac{n-1}{2}}g_{1}^{\frac{n+1}{2}})\cdots (g_{z}g_{z}^{n-1}\cdots g_{z}^{\frac{n-1}{2}}g_{z}^{\frac{n+1}{2}}) $$ where $z=\vert G\vert$, you've written out far more than the product of just the elements of $G$ because a quick sanity check will tell you that $e,g_1,\dots,g_z$, without all the other elements you've included, accounts for $z + 1 > |G|$ many elements, so this is ostensibly not the product you are hoping it is.
Since you've already got the idea, pair an element with its inverse, try this: $$ \big(\prod_{g\in G}g\big)^2 = \prod_{g\in G}gg^{-1} = e. $$ (Fill in the missing steps for completeness if you like; they are small.) Hence what can you say about the order of the element $\prod_{g\in G}g$? Finish with Lagrange's theorem.