A Kuratowski closure operator $f:\mathcal P(X)\to\mathcal P(X)$ has the property that there is a (unique) topology $\tau$ on $X$ such that $f$ is the closure operator for that topology, i.e., for every subset $U$ of $X$, $f(U)$ is the $\tau$-closure of $U$.
A "user2520938 closure operator" $f:\mathcal P(X)\to\mathcal P(X)$ in general will not have that nice property. For example, if $(X,\tau)$ is a topological space, and if I define $f:\mathcal P(X)\to\mathcal P(X)$ by setting $f(U)=U$ if $U$ is $\tau$-closed and $f(U)=\emptyset$ otherwise, then $f$ is a "user2520938 closure operator".
In other words, your conditions, besides being more complicated, are insufficient to characterize topological closure operators.
Suppose we have a co-topology $\kappa$ and we define
$$c(A) = \bigcap \{C \in \kappa: A \subseteq C\}$$
so that $c(A) \in \kappa$ ($\kappa$ is closed under intersections) and $A \subseteq c(A)$. The intersection is non-void because $X \in \kappa$ is always one of the sets in the intersection.
We want to see that $c(A \cup B) = c(A) \cup c(B)$.
It's already clear from the definition that $A \subseteq B$ implies $c(A) \subseteq c(B)$. (the sets from $\kappa$ that contain $B$ also contain $A$, so $c(A)$ is the intersection of possibly more sets than $c(B)$ hence smaller).
So $A \subseteq c(A), B \subseteq c(B)$ and so $A \cup B \subseteq c(A) \cup c(B)$. The right hand side is in $\kappa$ as this is closed under finite unions, and as it contains $A \cup B$, $c(A) \cup c(B)$ is one of the sets being intersected in the definition of $c(A \cup B)$ and so
$$c(A \cup B) \subseteq c(A) \cup c(B)$$
On the other hand $A,B \subseteq A \cup B$ so $c(A), c(B) \subseteq c(A \cup B)$
and it follows that
$$c(A) \cup c(B) \subseteq c(A \cup B)$$
Hence we have equality.
We also have that $c(A) = A$ iff $A \in \kappa$, to answer the final question affirmatively. $c(A) = A$ implies $A \in \kappa$ because $c(A) \in \kappa$ always. And if $A \in \kappa$, $A$ itself is in the intersecting family defining $c(A)$ so that $A \subseteq c(A) \subseteq A$ and $c(A) = A$.
Best Answer
If $s: \mathbb{N} \to X$ is a sequence in $X$, we can define that $s$ converges to $x \in X$ by
$$x \in \bigcap \{\operatorname{cl}(s[A]) : A \subseteq \mathbb{N} \text{ infinite }\}$$
which can be shown by considerations as William Elliott gave as well.