Interesting question, I never realised we could see a category in this way!
Before I answer your question, let me first point out where in MacLane's Categories for the Working Mathematician the relevant definitions can be found (in case anyone else is interested). Page 10 defines the sets $O$, $A$ and the product $A \times_O A$, together with how these sets with the necessary operations form a category. Earlier, in the introduction, on page 2 it is explained what a monoid is in terms of diagrams (with their usual meaning, with respect to the cartesian product $\times$). Then on page 4 it is explained how we can replace the usual cartesian product $\times$ by a similar binary operation $\square$.
The question is then: given a category with $A$ and $O$ as its set of arrows and objects respectively, in the case where this operation $\square$ is given by $\times_O$, what is the neutral element for the monoid $(A, \circ)$?
For any such operation $\square$, the neutral element will be given by an arrow $1 \to A$, where $1$ is the unit of the operation $A \square A$. That is, for all objects $X$, we must have isomorphisms
$$
1 \square X \cong X \cong X \square 1.
$$
Note that if $\square$ is the cartesian product $\times$, then 1 is just a singleton and an arrow $1 \to A$ just amounts to an element of $A$. So this is why we usually think of it as just an element. For other operations than the cartesian product, the unit may be something more complex than just a singleton and we get a 'neutral element' that is no longer just an element.
One of these more complex cases is where $\square$ is $\times_O$. First let us determine the unit. I will denote the unit of this operation by $I$ to avoid confusion in notation. So we want some set $I$ such that for all $B \subseteq A$ ($\times_O$ is only defined on subsets of $A$), there are isomorphisms:
$$
I \times_O B \cong B \cong B \times_O I.
$$
It turns out that taking
$$
I = \{ id_a : a \in O \}
$$
will do the job (check this!).
Now the 'neutral element' (which is no longer just one element) will be given by an arrow $I \to A$. There is only one, very natural, choice here and that is the inclusion. You should check that this indeed makes the entire construction into a monoid. So the 'neutral element' is now actually a subset of $A$.