The result follows for arbitrary finite abelian groups from the $p$-group case.
Remember that a finite abelian group $G$ is the direct sum of its $p$-parts,
$$G = G(p_1)\oplus \cdots\oplus G(p_n),$$
where $p_1,\ldots,p_n$ are the distinct primes that divide $|G|$, and
$$G(q) = \{ a\in G\mid q^ma = 0 \text{ for some }m\geq 0\},\qquad q\text{ a prime.}$$
If $a\in G$ is of maximal order, then we can write $a=a_1+a_2+\cdots+a_n$, where $a_i\in G(p_i)$. Since $a$ is of maximal order in $G$, then $a_i$ is of maximal order in $G(p_i)$. By the $p$-group case, we can write $G(p_i) = \langle a_i\rangle\oplus H_i$ with $H_i\leq G(p_i)$. Then $H_1+\cdots+H_n$ is a subgroup of $G$, it is the internal direct sum of the $H_i$, and since $G(p_i) =\langle a_i\rangle\oplus H_i$, then
$$\begin{align*}
G &= G(p_1)\oplus \cdots \oplus G(p_n)\\
&= (\langle a_1\rangle\oplus H_1) \oplus \cdots \oplus (\langle a_n\rangle \oplus H_n)\\
&= (\langle a_1\rangle\oplus\cdots \oplus\langle a_n\rangle) \oplus (H_1\oplus\cdots\oplus H_n).
\end{align*}$$
To finish off, note that $\langle a_1\rangle\oplus\cdots\oplus \langle a_n\rangle = \langle a\rangle$ (e.g., by the Chinese Remainder Theorem).
You are assuming we have an abelian tower for the finite group $\;G\;$ :
$$(**)\;\;\;1=G_m\lhd G_{m-1}\lhd\ldots\lhd G_1\lhd G_0:=G\;,\;\;s.t.\;\;G_i/G_{i+1}\;\;\text{abelian}\;\;\forall\,1=0,1,...,m-1 $$
The above means in particular that $\;G_{m-1}\cong G_{m-1}/G_m\;$ is abelian, so by the part marked in red in the proof, there's a cyclic refinement of it:
$$1= A_0\lhd A_1\lhd\ldots\lhd A_{m_1}:=G_{m-1}\;,\;\;A_k/A_{k+1}\;\;\text{cyclic}$$
But also $\;G_{m-2}/G_{m-1}\;$ is abelian, so again by the red part we've a cyclic refinement
$$G_{m-1}=:B_0\lhd B_1\lhd\ldots\lhd B_{m_2}:=G_{m-2}\;,\;\;B_i/B_{i+1}\;\;\text{cyclic}$$
Observe now that the subrefinement ("sub" because it is a refinement of part of the original tower)
$$1=G_m:=A_0\lhd A_1\lhd\ldots\lhd A_{m_1}=G_{m_1}=B_0\lhd B_1\lhd\ldots\lhd B_{m_2}=G_{m_2}$$
is cyclic! Well, go on like this inductively up through the whole first, original tower (**) ...
Best Answer
The theorem is saying that if $G/Z(G)$ is cyclic then $G$ is abelian so $G/Z(G)$ has one element. Putting these together: if $G/Z(G)$ is cyclic then it has one element.
The hypothesis here is not that $G$ is abelian. That is a consequence of the assertion that $G/Z(G)$ is cyclic. The interesting result here is that $G/Z(G)$ can never be a nontrivial cyclic group, like $\mathbb{Z}/12\mathbb{Z}$ or $\mathbb{Z}/2013\mathbb{Z}$.
The title of the theorem is not "factor group of an abelian group by its center is cyclic". It is "if the factor by the center is cyclic, then the group was abelian (so the factor was trivial)."