Valuations with the same valuation ring

Your proof is fine and it is indeed sufficient to consider only the multiplicative structure on $k^\times$, i.e., you need not be worried that you did not use the min-property for addition at all.

Here's the same result from a different perspective:

Any group homomorphism $f\colon G\to H$ from a group $G$ to an ordered group $H$ straightforwardly induces an order on $G/\ker f$ by us declaring those elements of $G/\ker f$ positive/negative that are mapped to positive/negative elements of $H$. With this, the induced monomorphism $\overline f\colon G/\ker f\to H$ is a monomorphism of ordered groups and hence an isomorphism of ordered groups from $G/\ker f$ to the image $\operatorname{im}\overline f$. In particular, if $f$ is onto, we obtain a isomorphism between ordered groups.

Now in the OP situation, we are given that the two valuations $v$ and $v'$ have the same kernel $$\begin{align}N:=\ker v&=\{\,a\in k^\times\mid v(a)\ge0\land v(a^{-1})\ge 0\,\}\\&=\{\,a\in k^\times\mid v'(a)\ge0\land v'(a^{-1})\ge 0\,\}=\ker v'\end{align}$$ and that they induce the same order on the quotient $k^\times/N$. From the above, this is isomorphic as ordered group to both $H$ and $H'$, i.e., $H$ and $H'$ are isomorphic as ordered group, as was to be shown.