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"
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 $R^*=R\setminus\{0\}$).