Let $(H,+)$ be any abelian group, and let $\mathbb{Z}\leq G\leq \mathbb{Q}$ a subgroup of the rational numbers with respect to addition. Let $\varphi:G\rightarrow H$ be a map with the following properties
(1) The restriction of $\varphi$ to $\mathbb{Z}$ is a homomorphism. (i.e. $\varphi(n) = n\varphi(1)$ and $\varphi(0)=0_H$)
(2) For every rational number $q\in G$, we can find $a,b\in\mathbb{Z}$ such that $q=\frac{a}{b}$ and $\varphi(q)^b = \varphi(a)$.
Is it necessarily true that $\varphi$ is an (additive) homomorphism. That is for all $q,p\in G$ we have $\varphi(p+q) = \varphi(p) + \varphi(q)$?
You should have written $b\varphi(q)$ instead of $\varphi(q)^b$ in (2) as you write groups additively.
The answer is negative (even if we replace "we can find" with "for any").
Take $G=\{n/3 : n\in\mathbb{Z}\}$, $H=\mathbb{Z}\times(\mathbb{Z}/3\mathbb{Z})$ (the second "multiplier" is $\{0,1,2\}$ with addition modulo $3$) and $\varphi(n/3)=(n,d(n))$, where $d(n)=0$ if $3|n$ and $d(n)=1$ otherwise. Then $\varphi$ meets (1) and (2), but $$(2,1)=\varphi(2/3)\neq\varphi(1/3)+\varphi(1/3)=(2,2).$$