This isn't a complete answer, but I thought it would be useful to write it out in more detail than would fit in the space provided.
First, I'd like to observe that both of the equivalent definitions of group action admit generalisations:
The definition using a homomorphism $G \to \text{Aut}(X)$ generalises straightforwardly to the case when $G$ is a (discrete) group and $X$ a vector space, giving rise to the notion of linear representations of groups.
The definition using a map $G \times X \to X$ generalises straightforwardly to the case when $G$ is a topological group and $X$ a topological space, giving rise to the notion of continuous group actions on topological spaces.
Your question seems to be asking whether we can find a definition which would encompass both branches of generalisations. It's not an unreasonable question — after all, both of the examples above converge when we have a continuous linear representation of a topological group.
Let's try to formulate categorically the equivalence of the two definitions in $\textbf{Set}$, and see how things go when we try to change to a more general category. Firstly, we note that a small group is equivalently
- a category $\mathcal{G}$ enriched over $\textbf{Set}$ with one object and all arrows invertible, and
- a set $G$ equipped with maps $m : G \times G \to G$, $e : 1 \to G$, $i : G \to G$ satisfying the group axioms.
The connection between the two definitions is expressed by the equation $\mathcal{G}(*, *) = G$, where $*$ is the unique object in $\mathcal{G}$. Using the first definition, a group action of $G$ is simply any functor $\mathscr{F} : \mathcal{G} \to \textbf{Set}$. Focusing on the hom-sets, we see that we have a monoid homomorphism $G \to \textbf{Set}(X, X)$, where $X = \mathscr{F}(*)$, and since the domain is a group, the codomain must lie in $\text{Aut}(X) \subseteq \textbf{Set}(X, X)$. Thus we have the first definition of group action.
Now, we recall that $\textbf{Set}$ is a cartesian closed category (indeed, a topos), so in particular we have the exponential objects $Y^X$, which are defined by following the universal property: $\text{Hom}(Z \times X, Y) \cong \text{Hom}(Z, Y^X)$ naturally in $Z$ and $Y$. Thus, we may identify the map $G \to \textbf{Set}(X, X)$ with a map $G \times X \to X$, and translating the homomorphism axioms through this identification gives the second definition of group action.
Consider a group object $G$ in a cartesian monoidal category $(\mathcal{C}, \times, 1)$. This may be viewed as a category $\mathcal{G}$ enriched over $\mathcal{C}$. Then, an action of $G$ on an object $X$ in another category $\mathcal{D}$ enriched over $\mathcal{C}$ is simply a $\mathcal{C}$-enriched functor $\mathcal{G} \to \mathcal{D}$. If $\mathcal{D}$ is such that there is a $\mathcal{C}$-enriched "forgetful" functor $U : \mathcal{D} \to \mathcal{C}$ and $\mathcal{C}$ is a cartesian closed category, we may do the same trick as before and obtain an arrow $G \times X \to X$ in $\mathcal{C}$.
In particular, if $\mathcal{C}$ is a cartesian closed category, it is enriched over itself. Indeed, consider the counit $\epsilon_{Z,X} : Z^X \times X \to Z$ of the product-exponential adjunction. If we compose with $\text{id} \times \epsilon_{X,Y} : Z^X \times X^Y \times Y \to Z^X \times X$, we get a map $Z^X \times X^Y \times Y \to Z$, which we may take transpose to obtain a map $Z^X \times X^Y \to Z^Y$, which is the composition of arrows. Thus we may recover the notion of continuous group actions at least in the case where both the group and the space being acted on are compactly-generated Hausdorff spaces...
What the definition says is that given a functor $\mathcal F \colon \mathbf C \to \mathbf{Set}$ a universal element is an element $e$ of a $\mathcal F(r)$, for some $r \in \mathbf C$, such that every other element $x$ of another set $\mathcal F(c)$, for a $c \in \mathbf C$, it can be viewed as image of a unique function of type $\mathcal F(r) \to \mathcal F(c)$. This function must be of course the image of a morphism of $\mathbf C$ via the functor $\mathcal F$.
Universal element play an important role in mathematics, primarily because $\mathbf{Set}$-based functors are really important in maths.
First of all let's make a point clear: universal arrows and universal elements are essentially the same thing, I'll try to explain way.
The fundamental idea is that there's a bijection between a set $X$ and the set $\mathbf{Set}(\bullet,X)$, where $\bullet$ denote the singleton set $\{\emptyset\}$: we can identify every element $e \in X$ with the (unique) morphism $ \tilde e \colon \bullet \to X$ such that $\tilde e(\emptyset)=e$.
Let $\mathcal F \colon \mathbf C \to \mathbf{Set}$ be a functor, then a pair $\langle r \in \mathbf C, e \in \mathcal F(r) \rangle$ is universal element if and only if the pair $\langle r\in \mathbf C, \tilde e \colon \bullet \to \mathcal F(r) \rangle$ is a universal arrow from the object $\bullet$ to the functor $\mathcal F$.
Similarly given a functor $\mathcal F \colon \mathbf C \to \mathbf D$ a pair $\langle r \in \mathbf C, \tilde e \in \mathbf D(d, \mathcal F(r))\rangle$ is a universal arrow from an object $d \in \mathbf D$ to $\mathcal F$ if and only if it is also a universal element for the functor $\mathbf D(d,\mathcal F(-)) \colon \mathbf C \to \mathbf{Set}$.
Remind that categories are usually considered as object living inside the category $\mathbf {Set}$ so universal elements can be thought as a way to talk about universals of a category $\mathbf C$ in $\mathbf{Set}$-terms (externally to the category $\mathbf C$) while universal arrows are the way to think universals in the term of category $\mathbf C$ itself (internally to the category $\mathbf C$).
Both these point of view are actually useful: the first enables us to work in set-theoretic terms and so to work in a familiar framework, the second enable to work internally to the category we are considering, and this is really important from a logical point of view.
About the request for your application what are you looking for is a pair $\langle \Omega \in \mathbf{Set}, \Theta \subseteq \Omega \rangle$ such that for every other pair $\langle X \in \mathbf {Set}, Y \subseteq X \rangle$ there exists a unique morphism
$$f \in \mathbf{Set}^\text{op}(\Omega,X) = \mathbf{Set}(X,\Omega)$$
such that
$$\mathcal P (f)(\Theta)=f^{-1}(\Theta)=Y\ .$$
If we take
$$\Omega = \mathcal P(\{\emptyset\})=\{\emptyset,\{\emptyset\}\}=\{0,1\}$$
what we get is the $$\mathbf{Set}^\text{op}(\Omega,X)=\mathbf{Set}(X,\Omega)$$
is the set of characteristic functions of $X$.
For each $Y \subseteq X$ there's a unique $f \colon X \to \Omega$ such that $f^{-1}(1) = Y$, but this means exactly that for every pair $\langle X \in \mathbf{Set}, Y \in \mathcal P(X)\rangle$ there exists always a unique $f \in \mathbf{Set}^\text{op}(\Omega,X)$ such that $\mathcal P(f)(1)=Y$, and thus $\langle \Omega,1\rangle$ is a universal element for the functor $\mathcal P$.
I hope this answer may help you.
(Edit:) I want to add something else. Universal elements of a functor $\mathcal F \colon \mathbf C^\text{op} \to \mathbf{Set}$ are important also for another reason: via yoneda-lemma universal elements are those objects corresponding to natural isomorphism from functor $\mathbf C(-,\Omega)$ to the functor $\mathcal F$.
So a functor $\mathcal F$ has a universal element if and only if it is representable. This is important because representable functors actions can be represented in term of the category $\mathbf C$ itself: for instance for every morphism $f \in \mathbf C(c,c')$ we can represent the function $\mathcal F(f) \colon \mathcal F(c) \to \mathcal F(c')$ as the function
$$ g \in \mathbf C(c',\Omega) \mapsto g \circ f \in \mathbf C(c,\Omega)$$
identifying sets $\mathcal F(c)$ and $\mathcal F(c')$ with sets $\mathbf C(c, \Omega)$ and $\mathbf C(c',\Omega)$.
Best Answer
Defining $S \times G \to S$ by $(s,g) \mapsto s \cdot g := F(g)(s)$, the axiom $(s \cdot g) \cdot h = s \cdot (gh)$ follows from $F(gh) = F(h) \circ F(g)$, since $$s \cdot (gh) = F(gh)(s) = \big( F(h) \circ F(g) \big)(s) = F(h) \big(F(g)(s) \big) = F(h)(s \cdot g) = (s \cdot g) \cdot h.$$