First of all, the set of all implicit operations on finite semigroups (on a given alphabet $A$) can be identified with the free profinite monoid. This monoid is the completion of the free monoid $A^*$ for the profinite metric $d$, defined as follows.
A monoid morphism $f:A^* \rightarrow M$ separates two words $u$ and $v$ of $A^*$ if $f(u) \not= f(v)$. By extension, we say that a monoid $M$ separates
two words if there is a morphism from $A^*$ onto $M$ that separates them.
Given two words $u, v \in A^*$, we now set
\begin{align*}
r(u,v) &= \min \left\{|M| \mid \text{$M$ is a monoid
that separates $u$ and $v$} \right\} \\
d(u,v) &= 2^{-r(u,v)}
\end{align*}
with the usual conventions $\min \emptyset = +\infty$ and $2^{-\infty} = 0$.
Then $d$ is a metric (it is even an ultrametric) and
the completion $\widehat{A^*}$ of the metric space $(A^*, d)$ is a compact topological monoid. Indeed, one can show that the product is uniformly continuous on $A^*$ and hence extends by continuity to $\widehat{A^*}$.
The elements of $\widehat{A^*}$ are called profinite words. Given a word, or even a profinite word, one can show that the sequence $u^{n!}$ is Cauchy and hence converges to an element $u^\omega$. Moreover $u^\omega$ is an idempotent of $\widehat{A^*}$.
Since $A^*$ is countable, there are also countably many profinite words involving the $(-)^\omega$ operator. But as $\widehat{A^*}$ is uncountable, there are way more profinite words, but it is not so easy to give examples. Since $\widehat{A^*}$ is compact, one can extract a converging sequence from any sequence of words and its limit will be a profinite word. However, this does not give an explicit example.
First example. For each prime $p$, the sequence $u^{p^{n!}}$ is Cauchy and hence converges to an element denoted $u^{p^\omega}$.
Second example. Let us fix a total order on the alphabet $A$ and let $u_0, u_1, \ldots$ be the ordered sequence of all words of $A^*$ in the induced shortlex
order. For instance, if $A = \{a, b \}$ with $a < b$, the first
elements of this sequence would be
$$
1, a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab, bba, bbb,
aaaa, \ldots
$$
It is proved in [1, 2] that the sequence of words $(v_n)_{n \geq 0}$ defined by
$$
v_0 = u_0 ,\ v_{n+1} = (v_nu_{n+1}v_n)^{(n+1)!}
$$
converges to a profinite word $\rho_A$, which is idempotent and belongs to the minimal ideal of $\widehat{A^*}$.
[1] J. Almeida and M. V. Volkov, Profinite identities for finite semigroups
whose subgroups belong to a given pseudovariety, J. Algebra Appl.
2,2 (2003), 137--163.
[2] N. R. Reilly and S. Zhang, Decomposition of the lattice of pseudovarieties of finite semigroups induced by bands, Algebra Universalis 44, 3-4 (2000), 217--239.
