An abelian group has Prüfer rank at most one iff it is isomorphic to a subgroup of either $\Bbb Q$ or $\Bbb Q/\Bbb Z$. This is Exercise 4.2.6 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.
The Details:
On page 98 to 99, ibid., we have

Let $G$ be an abelian group and let $S$ be a nonempty subset of $G$. Then  $S$ is called linearly independent, or simply independent, if $0\notin S$ and, given distinct elements $s_1,\dots, s_r$ of $S$ and integers $m_1,\dots, m_r$, the relation $m_1s_1+\dots+m_rs_r=0$ implies $m_is_i=0$ for all $i$.

. . . and . . .

If $p$ is prime and $G$ is an abelian group, the $p$-rank of $G$
$$r_p(G)$$
is defined as the cardinality of a maximal independent subset of elements of $p$-power order. Similarly the $0$-rank or torsion-free rank
$$r_0(G)$$
is the cardinality of a maximal independent subset of elements of infinite order. Also important is the Prüfer rank, often just called the rank of $G$,
$$r(G)=r_0(G)+\max_{p} r_p(G).$$

(See here for how the $\sup$ instead of the $\max$ is more appropriate. I don't think it effects the exercise is question though.)
A previous question might be helpful here. It's Exercise 4.2.2:
If $G$ is abelian, show the Prüfer rank $r(G)$ is finite iff $\max\{ d(H)\}$ is finite, for $d(H)$ the min number of generators of f.g. $H\le G$
That exercise has a second part; namely, showing that $r(G)$ is finite implies $r(G)=\max\{ d(H)\}$.
The Question:

Prove that an abelian group has rank $\le 1$ if and only if it is isomorphic with a subgroup of $\Bbb Q$ or $\Bbb Q/\Bbb Z$.

Thoughts:
Let $A$ be an abelian group.

For the sufficiency, if $A\le\Bbb Q/\Bbb Z$, then $A$ clearly has no elements of infinite order, so $r_0(A)=0$. We need to show, then, in this case, that
$$\max_{p} r_p(A)\le 1.$$
Here, I am stuck.
I don't know what to make of when $A\le \Bbb Q$.

The necessity has me stumped too.
If $r(A)=1$ while $\max_{p} r_p(A)=0$, then $r_0(A)=1$, then $A$ would have all its nontrivial elements of infinite order. That's as far as I got. I'm guessing that, here, $A\cong \Bbb Z$
As I said above, I think I need to use Exercise 4.2.2, but I'm not sure how.

This appears to be a question I could answer myself with more time; however,  I have given it a few days and I would like to move on. The type of answer I'm looking for is a full solution but I would be happy with strong hints.
Please help :)
 A: Sufficiency.
If $A\leq\mathbb{Q}$, and $a,b\in A$ are nontrivial, then there exist nonzero integers such that $na=mb$. Thus, $na-mb=0$ with $na\neq 0$, $mb\neq 0$. This shows that the rank of $A$ is at most $1$ (exactly one if $A$ is nontrivial).
Note that $\mathbb{Q}/\mathbb{Z}\cong\bigoplus_{p\text{ prime}}\mathbb{Z}_{p^{\infty}}$, where $\mathbb{Z}_{p^{\infty}}$ is the Prüfer $p$-group. Indeed, we can invoke the structure theorem of divisible groups for this; alternatively and more directly, the Prüfer $p$-groups embed into $\mathbb{Q}/\mathbb{Z}$ with images that intersect trivially, giving an embedding from the direct sum into $\mathbb{Q}/\mathbb{Z}$. To show the map is surjective, consider an arbitrary $\frac{a}{b}\in\mathbb{Q}$ with $\gcd(a,b)=1$; factoring $b$ into prime powers, $b=p_1^{\alpha_1}\cdots p_r^{\alpha^r}$, we want to express the fraction as
$$\frac{a}{b} +\mathbb{Z} = \frac{s_1}{p_1^{\alpha_1}} + \cdots + \frac{s_r}{p_r^{\alpha_r}} + \mathbb{Z},$$
and letting $b_i = b/p_i^{\alpha_i}$, this becomes
$$\frac{a}{b} + \mathbb{Z} = \frac{s_1b_1+\cdots + s_rb_r}{b} + \mathbb{Z}.$$
Since $\gcd(b_1,\ldots,b_r)=1$, we can express $a$ as an integer linear combination of $b_1,\ldots,b_r$; and by adding a suitable multiple of $b=\mathrm{lcm}(b_1,\ldots,b_r)$, we may choose $s_1,\ldots,s_r$ to be nonnegative. Thus, the embedding is surjective, as desired.
Given this, we see that the $p$-rank of any subgroup of $\mathbb{Q}/\mathbb{Z}$ is at most one for each prime $p$ (since the Prüfer groups are quasicyclic, so not two $p$-power elements can be independent). Thus, the rank of a group $A\leq\mathbb{Q}/\mathbb{Z}$ is at most $1$.
Necessity.
(I'm using multiplicative notation for $A$)
Note that if $H\leq A$, then the Prüfer rank of $H$ is less than or equal to the Prüfer rank of $A$, since any witnesses to the torsionfree or $p$-rank for $H$ will also work for $A$. Thus, every subgroup of $A$ also has rank at most $1$. In particular, every finitely generated subgroup of $A$ has Prüfer rank at most $1$.
If $A_{\rm tor}$  is nontrivial but not all of $A$, then we can find an $a\in A$ that maps to a nontrivial element of $A/A_{\rm tor}$, and an element of order $p$, to show that the Prüfer rank of $A$ is at least two. So either $A$ is torsion or $A$ is torsionfree.
If $A$ is torsion, then each $p$-part is cyclic or quasicyclic. Indeed, any finitely generated subgroup of the $p$-part of $A$ has rank at most $1$, and hence is cyclic. Thus, the $p$-part of $A$ is isomorphic to a cyclic group or to a copy of $\mathbb{Z}_{p^{\infty}}$, and in either case it can be embedded in $\mathbb{Z}_{p^{\infty}}$. This holds for each $p$, and since $A$ is the direct sum of its $p$-parts, we conclude that $A$ is isomorphic to a subgroup of $\oplus_p\mathbb{Z}_{p^{\infty}}\cong\mathbb{Q}/\mathbb{Z}$.
If $A$ is torsionfree, let $a\in A$ be nontrivial; we define an embedding $\phi\colon A\to \mathbb{Q}$. We start by letting $\phi(a)=1$.
Given $b\in A$, we know there exist nonzero integers $n$ and $m$ such that $a^n=b^m$. Define $\phi(b) = \frac{n}{m}$.
This is well defined: if $a^r=b^s$ also holds, then $a^{ns} = b^{ms} = a^{rm}$, and since $A$ is torsionfree it follows that $ns=rm$, and therefore that $\frac{n}{m} = \frac{r}{s}$.
Also, $\phi$ is a homomorphism: if $b,c\in A$, with $a^n=b^m$, $a^r=c^s$, then $\phi(b)=\frac{n}{m}$, $\phi(c)=\frac{r}{s}$, and
$$a^{ns+mr} = (a^n)^s(a^r)^m = b^{ms}c^{sm} = (bc)^{ms}$$
so
$$\phi(bc) = \frac{ns+mr}{ms} = \frac{n}{m}+\frac{r}{s} = \phi(b)+\phi(c).$$
Finally, $\phi$ is one-to-one, since $n$ and $m$ are nonzero when $b$ is nontrivial. Thus, $A$ is isomorphic to a subgroup of $\mathbb{Q}$, as desired.
