Saying $a \in b$ in Category Theory
Any "categorical definition" should be invariant under equivalences of categories. If $p$ is any set with only one element $\star$, then $X \mapsto X \cdot p$ is an auto-equivalence of $\mathsf{Set}$. Here, we have $X \cdot p := \coprod_{x \in X} p := \bigcup_{x \in X} p \times \{x\} = \{(\star,x) : x \in X\}$. If $X$ is empty, then $X \cdot p$ is also empty. But the elements of $X \cdot p$ are never empty. Thus, if $\emptyset \in X$, then $\emptyset \cdot p = \emptyset \notin X \cdot p$.
Therefore, there is no categorical definition of $\in$. This is not really a coincidence or even a defect of category theory. Instead, category theory systematically replaces membership by morphisms. See also ETCS.
If your category has a terminal object and power objects, then for each object $X$ there is a distinguished element $\{ X \} : 1 \to \mathcal{P} X$.
If you had a monomorphism $\mathcal{P} X \to Y$, then one might opt to use it to interpret $\mathcal{P}X$ as a subobject of $Y$. In this context, it would make sense to call $X$ an element of $Y$. But it only makes sense in this context. We could extend it to subobjects $Z \subseteq Y$ and say that $X \in Z$ if and only if our chosen morphism $\mathcal{P} X \to Y$ factors through a representative of the subobject $Z$.
This might be more meaningful if you equipped your category with extra structure: a distinguished class of monomorphisms that you call "$\subseteq$". Presumably, you'd want the objects together with all of the distinguished morphisms to form a poset, and probably other features. With this structure in place, we might now define $X \in Z$ to mean that there exists a commutative diagram
$$ \begin{matrix} 1 & \to & Z \\ \downarrow & & {\small|}\!\cap \\ \mathcal{P}X &\subseteq& Y \end{matrix}$$
where the left arrow is $\{ X \}$.
That said, it would be a very unusual situation for any of this to be of use. Other notions of element find much more utility, such as:
- The notion of a (global) element $1 \to X$
- The notion of a generalized element: any morphism at all with codomain $X$
- Restrictions of which objects can be used as the domain are sometimes useful
- The relation $\in$ related to power objects
However, if one insists on expressing $\in$, it is possible (and sometimes it is indeed done), in exactly the same manner as you described: an element of an object $x$ could be defined as a morphism $1\to x$, just as in $\Bbb{Set}$.
But in general, what these common source object, "$1$" would be? The terminal object? For example in the category of groups, these $1\to G$ homomorphisms are all trivial. On the other hand, the elements of $G$ are represented by $\Bbb Z\to G$ homomorphisms (identifying the element as the image of $1\in\Bbb Z$). In general universal algebra, the free algebra on $1$ generator will play the role of the common source "$1$".
This way formulas like $u\in x$ are expressible, however, as Martin commented below, while $x$ is an object, $u$ is an other kind of entity (an arrow to $x$). For the wished interpretation where $u$ is also an object, a canonical (or at least, fixed) functor $f:C\to (1\downarrow C)$ is indeed needed, with the predescribed $1$ object, (but it usually makes no sense).