Okay, based on the extensive discussion in comments, it seems that you want to consider the following:
Suppose that we have a well-defined collection of sets, which we want to make into a category by letting them be the objects, taking the collection of morphisms to be all set-theoretic functions between the two sets, and using regular composition and domain/codomain identifications. Can we prove that under these circumstances, for us to have a category then the categorical identity arrow must be the identity function for the set?
The key is that you have enough functions to "separate points". Given any $a,b\in A$, $a\neq b$, there exists a function $g\colon A\to A$ such that $g(a)\neq g(b)$. For example, define $g$ to be the function that maps $b$ to $a$, and maps everything else to $b$. (Compare this with the example I gave in the comments, where this does not hold).
So, fix a set $A$, and suppose that $f\colon A\to A$ is the arrow that satisfies the identity conditions (for all objects $B$ and $C$, and all arrows $g\colon A\to B$ and $h\colon C\to A$, $gf = g$ and $fh = h$).
Pick any $a$ and $b$ in $A$, $a\neq b$. Let $g$ be a function with $g(a)\neq g(b)$; then $gf(a) = g(a)$, so it follows that $f(a)\neq b$. This holds for every $b\in A-\{a\}$, so the only possibility is that $f(a)=a$. This holds for all $a\in A$, so $f$ must be the identity map.
You can generalize this to any set-based category in which you can either separate points, or "hit" any point: if for every object $A$ and every elements $a,b\in A$ with $a\neq b$, there either exists an object $B$ and a morphism $h\in\mathcal{C}(A,B)$ such that $h(a)\neq h(b)$; or else there exists an object $C$, and a morphism $g\in\mathcal{C}(C,A)$ for which there exists an element $c\in C$ such that $g(c)=a$; then the identity morphism of $A$ must be the identity map of $A$.
Indeed, suppose that $f$ is the identity morphism, and let $a\in A$. For each $b\in A$, $b\neq a$, either we have $B$ and $h$ as above, and $hf(a) = h(a)$ implies that $f(a)\neq b$; or else there exists $C$, $c$ and $g$ as above with $g(c) = a$. Then $fg = g$ gives that $f(a)\neq b$ (since $f(g(c))=b$ and $g(c)=a$ implies $a=b$). Either way, you get that for all $b\in A$ with $b\neq a$, $f(a)\neq b$. So the only possibility left is that $f(a)=a$. This holds for all $a\in A$, so $f=1_A$.
Note. In a sense, the condition is both necessary and sufficient, though for silly reasons: if the condition is not met by $A$ and $a$, then the identity map of $A$ cannot be the identity morphism, simply because the identity map of $A$ satisfies the given conditions: for all $b\neq a$ you have $1_A(a)\neq 1_A(b)$, and $1_A(a)=a$.
Added. This argument applies to categories such as topological spaces (because you always have the map from the $1$-element topological space to your toplogical space mapping the unique point to $a$); pointed topological spaces (the discrete 2-element pointed topological space maps the non-distinguished point to your favorite point); groups (you have maps from the cyclic group to any group, mapping the generator to your element $a$); and others. It's hard to make it work with posets as categories, because posets as categories are not really set-based categories (the objects are not usually sets and arrows set-theoretic functions between them); you can model them as set-based categories, but then the result need not hold: the example I gave in the comments can be thought of as the totally ordered set with two elements, for example, and here you don't have $id_A = 1_A$.
In $C/A$, $1_A$ is a final object in the category - for every object $f:X\to A$ in $C/A$ there is a unique morphism $f\to 1_A$ - that is, $\hom(f,1_A)$ is always a singleton.
You are confusing $1_A$, with $1_{1_A}$. $1_A$ is an object of $C/A$. $1_{1_A}$ is the identity morphism for that object. That morphism is always trivial with composition, as are all identity morphisms.
It might be useful to consider the basic case of $C=\text{Set}$. $\text{Set}$ has final objects equal to the singleton sets.
Given a set $A=\{1,2\}$, it turns out that $\text{Set}/A$ is pretty much $\text{Set}\times\text{Set}$. The sets corresponding to $f:X\to A\in \text{Set}/A$ are $f^{-1}(1)$ and $f^{-1}(2)$. Now, if $C_1,C_2$ both have final objects, then $C_1\times C_2$ has a final object. So the final object in $\text{Set}^2$ is just two pairs of singleton sets. But it we reverse the operation, we see that corresponds to an element of $f:X\to A$ which is $1-1$ and onto.
For more general sets $A$, there is a sense in which $\text{Set}/A$ is equivalent to something we can write as $\text{Set}^A$. Again, we see that the final objects must be final at every "point" $a\in A$, and thus that the final objects of $\text{Set}^A$ are just lists of singletons indexed by $A$.
This feature of $\text{Set}$ is specific to that category, but I think it is instructive to see it in action. The finality of $1:A\to A$ is true in all $C/A$.
Best Answer
This category is not unique, but it is unique up to isomorphism. So this definition is ambiguous in the same totally harmless sense as saying “the trivial group is the group with one element.” Indeed, any two one-element sets $(X,Y)$ give an example of $\mathbf 1$ in a unique way, once one set is chosen to contain the object and the other, the morphism.