Let's call the subfield from the "weird" definition $F_C$, the field generated by $1_K$ as $F_I$, and the intersection of all subfields as $F_S$. Proof that these are all equal:
Since $1_k\in F_C$, we have $F_I\subseteq F_C$, and $F_S\subseteq F_I$ by definition.
If we ask for the minimum field containing $1_K$, then we need to include $1_K+1_K$, and $1_K+1_K+1_K$, etc. For all such elements, we also need their additive inverse to be in $F_I$. That means we can define $f':\mathbb{Z}\rightarrow F_I$ in the same way as you defined $f:\mathbb{Z}\rightarrow K$, and this is a ring homomorphism. In fact, $f'(n)=f(n)$ for all $n$. So $f'$ is injective if and only if $f$ is, meaning we can lift $\overline{f}'$ into $F_I$ into the same way. But since the action of $f'$ is the same as $f$, then $\overline{f}'(x)=\overline{f}(x)$ for all $x\in\mathbb{Q}$ (or $\mathbb{Z}/p\mathbb{Z}$), since the lifting $\overline{f}$ is unique. This means:
$$ F_C=\overline{f}(\mathbb{Q}) = \overline{f}'(\mathbb{Q})\subseteq F_I$$
(or replace $\mathbb{Q}$ with $\mathbb{Z}/p\mathbb{Z}$).
Finally, note that any subfield $K'$ of the field $K$ must contain $1_K$, so it must contain the field that $1_K$ generates. Thus, $F_I\subseteq K'$ for all subfields. This means $F_I\subseteq F_S$, the intersection of all subfields.
Intuitively, $1_K$ is going to generate an additive group, and this is going to look like $\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z}$. In the second case you're done because it will also be multiplicatively closed. In the first case you will also need the inverses of all those elements, which will look like fractions in $\mathbb{Q}$. Specifically, the field of fractions (the minimal field containing such a ring) will be isomorphic to $\mathbb{Q}$, the field of fractions of $\mathbb{Z}$.
Best Answer
Let $E$ and $F$ be fields. Suppose there exists an injective ring homomorphism $\phi:E \to F$. Then by isomorphism theorem, $\phi(E) \cong E/\ker(\phi) = E$. Thus, $\phi(E)$ is a subfield of $F$ that is isomorphic to $E$. We sometimes abuse notation a little bit by saying that $E$ is a subfield of $F$.
Note: if $E$ is a field, then any ring homomorphism $\psi: E \to R$ is either injective or the zero map. This is due to the fact that $\ker(E)$ is an ideal of $E$ and the only ideals of $E$ are $(0)$ and $E$.