Let $S$ be a finite semigroup. An element $a$ is said to be maximal if for any $b \in S$ with $\langle a \rangle \subseteq \langle b \rangle$ we have either $a = b$ or $\langle b \rangle = S$. Since $S$ is a finite semigroup so that $S$ contains finite number of maximal elements says $a_1, a_2, \cdots a_k$. Note that for any $a \in S$, we have $a \in \langle a_i \rangle$ for some $1 \leq i \leq k$. I want to prove that
Consider the following conditions on $S$:
(i) $S$ has no subgroup $C_p \times C_p$ for some prime $p$.
(ii) For $e, f \in E(S),\; ef = fe$ we have $f = e$, where $E(S)$ is the set of all idempotents of $S$.
(iii) For all $x,y\in S$, if $xy=yx$ then $x,y\in\langle a\rangle$ for some maximal element $a$.
Then (iii) holds if and only if (i) and (ii) hold.
I have proved one side, but I am stuck in converse part, how to proceed further. It may be required to add one more condition in (i), (ii) which are equivalent to (iii). Thanks in advance for giving me your precious time.
As stated I think this is not true. Consider the semigroup $S=\{0,1,2,3\}$ with operation $n\oplus m=(n+m)\,\text{mod}\,2$. Since $n\oplus n=0$, the only idempotent is $0$, so (ii) holds. Also any subgroup must be contained in $\{s\in S\mid s\oplus0=s\}=\{0,1\}$, so (i) holds. Also $S$ is abelian. The monogenic subsemigroups of $S$ are $$ \langle 0\rangle=\{0\}, $$ $$ \langle 1\rangle=\{0,1\}, $$ $$ \langle 2\rangle=\{0,2\}, $$ $$ \langle 3\rangle=\{0,1,3\}. $$ None of these contain $2$ and $3$, so (iii) fails.
More generally, consider the following construction. Let $L=\mathbb Z_{\geq0}\times\mathbb Z_{\geq0}$, which is an abelian semigroup under addition. Fix an abelian group $G=C_n\times C_m$ and a finite downward closed subset $I_0\subset L$; that is, $L\setminus I_0$ is an ideal in $L$. Let $I=I_0\setminus\{0\}$. We will define an operation on $S=I\sqcup G$.
Let $\pi:L\to G$ denote the natural homomorphism. Consider the semigroup $S_0=L\times\{0,1\}$ where the operation on $\{0,1\}$ is multiplication. This has an ideal $S_1=S_0\setminus\{((0,0),1)\}$. We have a surjection $\psi:S_1\to S$ given by $$ \psi(x,z)=\begin{cases} x&\text{if }x\in I\text{ and }z=1,\\ \pi(x)&\text{otherwise}. \end{cases} $$ The operation on $S_1$ descends to $S$ via $\psi$, making $S$ a semigroup.
Now $S$ is abelian and generated by two elements, namely $\psi((0,1),1)$ and $\psi((1,0),1)$. The only idempotent is $\psi((0,0),0)$, so statement (ii) holds. Statement (i) holds iff $n,m$ are coprime. Statement (iii) holds iff $n,m$ are coprime and $\{(1,0),(0,1)\}\not\subseteq I$.
Note that statements (i) and (ii) are local; if they hold for $S$, they hold for any subsemigroup of $S$. If we want to prove (iii) using local statements, it suffices to consider a subsemigroup generated by two commuting elements. Assuming (ii), such a semigroup will be a quotient of the above construction (with the quotient map injective on $G$). So it may be fruitful to consider the special case of the above construction.