I'm going to answer in terms of ZF set theory since that is what most of us need.
In ZF, there is no definition for a set. It is a primitive idea. All you have is the notion set and membership and that gives you equality of sets, and in turn equality of members.
Whether or not two elements of a set are equal is again a matter of set equality. You don't need anything extra to tell you when two elements are equal.
It certainly can't be part of the definition of a set, because an equivalence relation is essentially a special subset of $X\times X$, and if you haven't accepted what a set is yet, you shouldn't be discussing thing like subsets of $X\times X$. You will just be going around in circles.
You can grant $X$ an equivalence relation given by the partition of the set into singletons, so that you get the "identity relation", but it doesn't tell you anything new.
E.g., can we call equinumerous sets isomorphic?
Sure you can. In the category of sets, the isomorphisms are precisely the bijections. There are "isomorphic in the category of sets."
You don't need operations to do this... a category can be made up of non-algebraic objects. That is isomorphism and homomorphism are not algebraic-only concepts.
Thus, every set by the definition is an algebraic structure with the identity operation.
It would be more plausible to say that a set is an algebraic structure with no operations. I don't know if universal algebra accepts this empty case, but they may.
Let us suppose we are dealing with an arbitrary partially ordered set $(P, \leq)$. In the particular case of topological spaces, $P$ is some collection of subsets of $X$, the underlying space. We can consider $P$ as a category in a canonical way as follows: the set of objects is $P$, there is at most one arrow between each $x, y \in P$, and there is an arrow between $x$ and $y$ iff $x \leq y$.
A sieve on an object $x$ can be defined as a collection $S \subseteq \{(f, z) | f : z \to x\}$ which satisfies the property that for every $(f, z) \in S$ and every $g : w \to z$, we have $(f \circ g, w) \in S$.
When we are talking about a partially ordered set, the first component of $(f, z)$ where $f : z \to x$ doesn't add any information (other than the fact that $z \leq x$) since there is at most one $f : z \to x$. Thus, we can equivalently consider a sieve $S$ on $x$ to be a collection $S \subseteq \{z \in P : z \leq x\}$ s.t. for all $z \in S$, for all $w \leq z$, $w \in S$. This is what I will call a PO-sieve.
Given a sieve $S$ on $y$ and an arrow $f : x \to y$, we may define the $f^*(S) = \{(g, z)| g : z \to x$ and $f \circ g \in S\}$, a sieve on $y$.
Correspondingly, given a PO-sieve $S$ on $y$ and some $x \leq y$, we may define $S_x = \{z : z \leq x$ and $z \in S\}$, a sieve on $x$.
A Grothendieck Topology on a category $C$ is a mapping from each object $x \in C$ to a family $F_x$ of sieves on $x$ which satisfies several axioms.
Correspondingly, a PO-Grothendieck Topology on a poset $P$ must be a mapping from each element $x \in P$ to a family $F_x$ of PO-sieves which satisfies the corresponding axioms.
Axiom 1 of Grothendieck Topology: for every $x \in C$, we have $\{(f, z) : f : z \to x\} \in F_x$.
Corresponding Axiom 1 of PO-Grothendieck Topology: for every $x \in P$, we have $\{z : z \leq x\} \in F_x$.
Axiom 2 of Grothendieck Topology: for every $f : x \to y$, for every sieve $S \in F_y$, we have $f^*(S) \in F_x$.
Corresponding Axiom 2 of PO-Grothendieck Topology: for every $x \leq y$ and for every PO-sieve $S \in F_y$, we have $S_x \in F_x$.
Axiom 3 of Grothendieck Topology: suppose we have $S \in F_x$. And suppose we have a sieve $P$ on $x$ such that for all $(f, z) \in S$, $f^*(P) \in F_z$. Then $P \in F_x$.
Corresponding Axiom 3 of PO-Grothendieck Topology: suppose we have $S \in F_x$. And suppose we have a PO-sieve $P$ on $x$ s.t. for all $z \in S$, $P_z \in F_z$. Then $P \in F_x$.
How does this relate to Generalised Topological Spaces? Suppose given such a generalised space. The partially ordered set $P$ is the set of opens ordered by $\subseteq$. Suppose given some collection $C$ of open sets. Define $f(C) = \{U $ open$: \exists V \in C (U \subseteq V)\}$. Note that for every such $C$, $f(C)$ is a PO-sieve. Then given $U$ open, we may define $F_U = \{f(C) : C \in cov_X$ and $\bigcup\limits_{V \in C} V = U\}$.
Let us verify that this gives us a PO-Grothendieck topology.
Axiom 1: this follows from the fact that $\{U\} \in cov_X$ for all $U$ - that is, it follows from axiom A3.
Axiom 2: this follows from axiom A5.
Axiom 3: this follows from axiom A6.
Finally, we turn to your example of $\mathbb{N}$ with "opens" recursively enumerable sets and "coverings" recursive enumerations of recursively enumerable sets. Since this satisfies axioms A3, A5, and A6, it does form a PO-Grothendieck topology.
Best Answer
In my experience the usual context in which saturated sets appear in topology is the one metioned by Stefan H. in the comments: you have a map $f:X\to Y$, and you say that a set $A\subseteq X$ is saturated with respect to $f$ iff $A=f^{-1}\big[f[A]\big]$. More generally, if $\mathscr{P}$ is a partition of $X$, a set $A\subseteq X$ is saturated with respect to $\mathscr{P}$ iff $A=\bigcup\{P\in\mathscr{P}:P\cap A\ne\varnothing\}$. (This really is a generalization: in the case of the map $f$, the associated partition of $X$ is $\{f^{-1}[\{y\}]:y\in Y\}$, the set of fibres of the map $f$.)
If $\mathscr{P}$ is a partition of $X$ and $A$ is an arbitrary subset of $X$, there are two saturated sets naturally associated with $A$. One, which we might call the outer saturation of $A$, is $$\bigcup\{P\in\mathscr{P}:P\cap A\ne\varnothing\}\;;$$ if $\mathscr{P}$ is generated as above by a map $f:X\to Y$, the outer saturation of $A$ is $A=f^{-1}\big[f[A]\big]$. The other, which we might call the inner saturation of $A$, is $$\bigcup\{P\in\mathscr{P}:P\subseteq A\}\;;$$ it’s the complement of the outer saturation of $X\setminus A$, so if $\mathscr{P}$ is generated as above by a map $f:X\to Y$, the inner saturation of $A$ is $X\setminus f^{-1}\big[f[X\setminus A]\big]$.
My best guess without having seen the actual context is that by the saturation of a set $A$ they mean what I’ve called here the outer saturation of $A$.
Note: The terms outer saturation and inner saturation are not standard, so far as I know; I’m using them here for purposes of exposition.