Let $A$ be a subgroup of $\mathbb{Q}^2$, generated by $e:=(0,1)$ and $f_n:=(1/n!,\alpha_n)$, where $\alpha_n = \frac{1}{n} (\alpha_{n-1} + \nu_n)$, $\alpha_1 = 1$, and $\nu_n \in \mathbb{Z}$, to be specified later.
First of all, note that for each $n$ the vectors $e$ and $f_n$ generate a subgroup $A_n$ that contains $A_{n-1}$, and that $A = \bigcup_n A_n$, so it can be easily checked that $A \cap (0 \times \mathbb{Q})$ is the subgroup generated by $e$, isomorphic to $\mathbb{Z}$. Furthermore, it is the kernel of the map that calculates the first coordinate, and the image is exactly $\mathbb{Q}$.
Now my aim is to choose $\nu_n$ in such a way that no element of $A$ is divisible enough. Clearly this can be done in many ways. Fix an $n$ for a moment, and notice that for $x \in A_n$ to be divisible by a large prime $p>n$ its second coordinate must equal $\frac{p!}{n!} \alpha_p$ modulo $p$. So by choosing $\nu_p$ we may ensure that a fixed $x$ is not divisible by $p$. What remains is to enumerate them carefully and choose $\alpha_p$$\nu_p$ in such a way that for every $n$ and every $x \in A_n$ there exists at least one $p$ such that $x$ is not divisible by $p$. Thus we need an injective map $(n,x) \mapsto p$ subject to $p > n$. It is easy but messy to write down... Just to elaborate the whole process: we choose $\nu_n$ in their usual order, and each time we run into a distinguished prime $p$ that is responsible for some $(n,x), n < p$, we should act accordingly using our knowledge of the previous $\alpha$'s.