An abelian torsion group has a unique basic subgroup iff it is divisible or bounded. This is Exercise 4.3.14 of Robinson's  "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
The Details:

Let $p$ be prime. A $p$-group is a group all of whose elements have a $p$ power order.

On page 12 of Robinson's book,

A torsion group [. . .] is a group all of whose elements have finite order.

On page 94, ibid.,

An element $g$ of an abelian group $G$ is said to be divisible in $G$ by a positive integer $m$ if $g=mg_1$ for some $g_1$ in $G$. [. . .]
An abelian group $G$ is said to be divisible of each element is divisible by every positive integer.

On page 106, ibid.,

A subgroup $H$ of an abelian group $G$ is called pure if
$$nG\cap H=nH$$
for all integers $n\ge 0$.

On page 107, ibid.,

Let $G$ be an abelian torsion group. A subgroup $B$ is called a basic subgroup if it is pure in $G$, it is the direct sum of cyclic groups, and $G/B$ is divisible.

On page 108, ibid.,

An additively written group is called bounded if its elements have boundedly finite orders.

The Question:

(Kulikov) Prove that an abelian torsion group has a unique basic subgroup if and only if it is divisible or bounded. [Hint: Let $B$ be the unique basic subgroup of the $p$-group $G$. Write $B=\langle x\rangle \oplus B_1$ and show that $G=\langle x\rangle \oplus G_1$ for some $G_1$. If $a\in G_1$ and $\lvert a\rvert\le\lvert x\rvert$, prove that the assignments $x\mapsto xa$ and $g_1\mapsto g_1$, $(g_1\in G_1)$, determine an automorphism of $G$. Deduce that $a\in B$.]

Thoughts:
The hint, in essence, I think, suggests that I use Exercise 4.3.8. I asked about that exercise here earlier:

*

*An abelian $p$-group has a bounded basic subgroup iff it is the direct sum of a divisible group and bounded group

Let $G$ be an abelian torsion group.

$\Rightarrow$
Suppose $B$ is a unique basic subgroup of $G$.
What allows us to assume $G$ is a $p$-group, following the hint?

$\Leftarrow$
It seems natural to use the following logical equivalence:
$$((D\lor B)\to U)\leftrightarrow ((D\to U)\land (B\to U)).$$

*

*$(D\implies U)$
Suppose $G$ is divisible. Then for all $g\in G$ we have for all $m\in\Bbb N$ there exists an $h\in G$ with
$$g=mh.$$
What do I do next? I have nothing nontrivial to add.

*

*$(B\implies U)$
Let $G$ be bounded. Then each $g\in G$ has boundedly finite order.
This is a stronger assertion than saying $G$ is a torsion group.
Again, I have nothing nontrivial to add.

I have skipped a few exercises from Section 4.3 of the book. This is primarily because I have spent too much time on the set; I forgot too much of what came before and I began to lose patience in myself. I chose this exercise because it seems interesting and I have experience with a similar problem (see above).
If I had more time, I think I could solve this myself. There's a lot going on in my life at the moment, though, so I cannot devote too much to one exercise, especially considering that the exercise is not marked as being referred to later on in the text.

Please help :)
 A: This is a CW answer adapting the comments by @DavidA.Craven above.
For bounded groups $G$, if $B\le G$ and $G/B$ is divisible (and bounded), then $G/B=1$. Thus $B=G$ and hence is unique.
Every quotient of a divisible group is divisible (if $x=n\cdot y$ then $x+H=n\cdot(y+H)$), so if $G$ is torsion divisible, we need that it has no non-trivial basic subgroups $B$. But torsion divisible groups are direct sums of quasicyclic groups (4.1.5), and pure subgroups are just direct summands by Exercise 4.3.3. Quasicyclic subgroups are not sums of cyclic subgroups, so $B=1.$
Finally, I know Sylow's theorem is still not allowed, but for torsion abelian groups the set of all elements of order a power of $p$ is a subgroup, and a torsion group is the direct sum of these over all $p$. I'm sure Robinson uses this all the time. If $G=\oplus_pG_p$ has a unique basic subgroup $B$, then $B=\oplus_pB_p$, and so the question is whether $B_p$ is basic in $G_p$. Clearly it is a sum of cyclic groups, $G_p/B_p$ is divisible, and so pure is all that is needed. This is also clear from the fact that $B$ is pure in $G$.
