Let $U(R)$ be the set of units in $R$ (the elements that have multiplicative inverses), and consider the subset $\{r\in U(R): f(r) = 1\}$. That is, the set of elements of $U(R)$ that are mapped to the multiplicative identity of $S$. This is analogous to the kernel of $f$, but clearly isn't a subring since if $f(r)=f(s)=1$ then $f(r+s) = f(r)+f(s)=1+1\ne1$. It seems to be a subgroup under multiplication though, since $f(rs) = f(r)f(s) = 1\cdot 1=1$, and $f(r^{-1}) = f(r)^{-1}=1^{-1}=1$. Is there a name or any other interesting properties of this subset (subgroup) of $U(R)$?
My apologies if this question has been asked before; since I don't have a name for it, I wouldn't know what to search for.
Every unital morphism between rings $f\colon R\to S$ induces a multiplicative group homomorphism, which we might denote $U(f)\colon U(R)\to U(S)$. Your set is $\mathrm{ker}(U(f))$.
Note that there is a functor from the category of rings with unity and unital morphisms (morphisms that must send the multiplicative identity to the multiplicative identity) to the category of groups, $U\colon \mathscr{R}ing^1\to\mathscr{G}roup$, defined by mapping a ring $R$ to its unit group $U(R)$, and a morphism $f\colon R\to S$ to the restriction $U(f)=f|_{U(R)}$. This is a functor, since $$\begin{align*} U(\mathrm{id}_{R}) &= \mathrm{id}_{U(R)}\\ U(g\circ f) &= (g\circ f)|_{U(R)} = g|_{U(S)}\circ f|_{U(R)} = U(g)\circ U(f). \end{align*}$$ Your set can also be seen as the image of the kernel under the functor.