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 $c(A) = A$.
Your definition of $\tau$ isn't right. By your definition $\tau$ is simply the powerset of $X$. Here is likely what you want:
$$\tau=\{X\setminus h(A)\mid A\subseteq X\}$$
Then $X\setminus h(\emptyset)=X\in\tau$ and $X\setminus h(X)=\emptyset\in\tau$.
If $A,B\in\tau$ then $A=X\setminus h(U)$ and $B=X\setminus h(V)$ for some $U,V\subseteq X$. Then
$$A\cap B=(X\setminus h(U))\cap(X\setminus h(V))=X\setminus(h(U)\cup h(V))=X\setminus h(U\cup V)$$
Therefore $A\cap B\in\tau$.
Finally if $\{A_{\alpha}\}_{\alpha\in I}$ is some collection of elements in $\tau$ then say that $A_{\alpha}=X\setminus h(U_{\alpha})$ for some $U_{\alpha}\subseteq X$. Then
$$\bigcup_{\alpha\in I}A_{\alpha}=\bigcup_{\alpha\in I}(X\setminus h(U_{\alpha}))=X\setminus\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
To finish this line we will need your third and second axioms.
$$\bigcap h(U_{\alpha})\subseteq h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
For each $\alpha\in I$ we have that $\bigcap_{\beta\in I}h(U_{\beta})\subseteq h(U_{\alpha})$
Thus by monotonicity and idempotence (axioms two and three) we have
$$h\left(\bigcap_{\beta\in I}h(U_{\beta})\right)\subseteq hh(U_{\alpha})=h(U_{\alpha})$$
for each $\alpha$. We then have that
$$h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)\subseteq\bigcap_{\alpha\in I}h(U_{\alpha})\subseteq h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
Establishing equality between the two sets. We then have that
$$X\setminus\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)=X\setminus h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
Therefore, the union over the family $\{A_{\alpha}\}_{\alpha\in I}$ is an element of $\tau$, establishing that $\tau$ is indeed a topology on $X$.
Best Answer
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.