If $\varphi: K \to K'$ is a map of Integral Domains st. $\varphi(xy)=\varphi(x) \varphi(y)$ and $\varphi(x+y)=\varphi(x) + \varphi(y)$. Then $\varphi(0)=0'$ as $\varphi(0)=\varphi(0+0)=\varphi(0)+\varphi(0) \implies \varphi(0)=0'$ by the cancellation law.
I see that we require $\varphi(1)=1'$ to avoid, for example, $\varphi(k)=0' \ \forall k \in K$
Why does the same not work to show that $\varphi(1)=1'$?
On
The example you gave shows why we can't prove that $\varphi(1)=1'$, because the $0$ map is an example of a homomorphism which doesn't map 1 to 1'.
If you try to do the same proof of $\varphi(0)=0'$ for $1$, you get $\varphi(1)=\varphi(1\times 1)=\varphi(1)^2$. Thus $\varphi(1)\times(\varphi(1)-1')=0''$. Thus $\varphi(1)=0$ or $\varphi(1)=1'$.
So you need to assume that $\varphi(1)\neq 0'$ in order to get $\varphi(1)=1'$.
Given a ring homomorphism of integral domains, i.e. a set-function $\varphi\colon R\to R'$ such that for all $x,y\in R$ we have $\varphi(x+y)=\varphi(x)+\varphi(y)$ and $\varphi(xy)=\varphi(x)\varphi(y)$, we find that there are two possibilities:
Indeed, mimicking the proof of $\varphi(0)=0$ we arrive at $$\varphi(1)=\varphi(1^2)=\varphi(1)^2\implies\varphi(1)[\varphi(1)-1]=0$$ As $R'$ is an integral domain we are left with the $\varphi(1)=0$ or $\varphi(1)=1$. Note that in the first case $\varphi(x)=\varphi(1x)=\varphi(1)\varphi(x)=0\varphi(x)=0$ for all $x\in R$.$~~~\square$
For this to work it is crucial that $R$ and $R'$ are integral domains (in fact, it is sufficient forcing $R'$ to be an integral domain). On the other hand, if $\varphi$ is surjective we can show that $\varphi(1)$ behaves like an identity in $R'$ and by the uniqueness of such elements we conclude that $\varphi(1)=1$. For further discussion, see this relevant post.