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$'s. 1 Let$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$in such a way that for every$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...