Suppose $S$ is a semigroup. Define $Add(S)$ as the set of all finite subsets of $S$, equipped with the operation of pairwise “addition” ($\forall A, B \in Add(S)$, $AB = \{ab| a \in A, b \in B\}$). It is not hard to see, that $Add(S)$ is closed under that operation and that that operation is associative, which makes $Add(S)$ a semigroup.
Does there exist such a non-trivial semigroup $S$ ($|S| > 1$), that $S \cong Add(S)$?
If such semigroup exists, it has to be infinite (if $S$ is finite, then $|Add(S)| = 2^{|S|} > |S|$ for any non-trivial $S$), and it has a proper subsemigroup isomorphic to itself (the subsemigroup of $Add(S)$ formed by $1$-element subsets) and a zero (empty subset). However, I failed to determine anything else about it.
If I am not mistaken, the answer is yes. Here is the idea.
Let $S$ be a semigroup. Then the semigroup of all finite subsets of $S$ is usually denoted by $\mathcal{P}_f(S)$. This operator $\mathcal{P}_f$ can be iterated by setting $\mathcal{P}_f^{(0)}(S) = S$ and $\mathcal{P}_f^{(n+1)}(S) = \mathcal{P}_f(\mathcal{P}_f^{(n)}(S))$ for all $n> 0$. Now, the map $x \to \{x\}$ defines an embedding from $S$ into $\mathcal{P}_f(S)$, yielding a directed system $$ S = \mathcal{P}_f^{(0)}(S) \to \mathcal{P}^{(1)}_f(S) \to \mathcal{P}^{(2)}_f(S) \to \dotsm $$ Since the category of semigroups has direct limits, the semigroup $\mathcal{P}_f^\omega(S) = \lim \mathcal{P}^{(n)}_f(S)$ is well-defined and by construction, $\mathcal{P}_f(\mathcal{P}_f^\omega(S))$ is isomorphic to $\mathcal{P}_f^\omega(S)$. This applies in particular to the trivial semigroup $S = \{1\}$.