Is there in fact a category of rings in which $ \mathbb{Z} $ is initial? If one does not require that the (set-)mappings preserve the multiplicative identity, then the null map is an endomorphism of $ \mathbb{Z} $ distinct from the identity
If it is indeed required, I understand that the composite
$ h: \mathbb{Z} \rightarrow \mathbb{Z_2} \rightarrow \mathbb{Z}$
is not in fact a morphism, since
$ h(2) = 0 \neq 2 = h(1) + h(1)$
(likewise for $ g : \mathbb{Z_2} \rightarrow \mathbb{F_1} \rightarrow \mathbb{Z_2} $, but in which case one can only conclude $ \mathbb{F_1} $ should not figure as an object to begin with)
so there is no category of "unital rings and mappings preserving both identities"
 A: I'm referencing the usual category of (commutative) unital rings, where the objects are the algebraic structures $(A,+,\cdot,0,1)$ such that $(A,+,0)$ is an abelian group, $(A,\cdot,1)$ is a (commutative) monoid and multiplication is distributive over sum. This definition includes the $0$ ring $(\{0\},+,\cdot,0,0)$. Every monoid has exactly one identity, therefore the indication of $0$ and $1$ is redundant, and I'll omit it. The morphisms $(A,+_A,\cdot_A)\to (B,+_B,\cdot_B)$ are the maps which are both group homomorphisms $(A,+_A)\to (B,+_B)$ and morphisms of monoids $(A,\cdot_A)\to (B,\cdot_B)$. Id est, they are the functions such that: \begin{cases}f(x+_A(-y))=f(x)+_B(-f(y))\\ f(x\cdot_Ay)=f(x)\cdot_Bf(y)\\ f(1_A)=1_B\end{cases}
In these two categories $\Bbb Z$ is the initial object and $0$ is the final object. So there really is no issue with there not being a map $\Bbb Z_2\to\Bbb Z$: the initial object just has maps from it to other objects, not the other way around.
On the other hand, a division ring is a non-zero ring $R$ such that $R^*=R\setminus\{0\}$ (or, in a funny equivalent way, a ring such that $R^*=R\setminus\{0\}$).
