Very interesting problem! I'd not seen it before, but from what I can make out, it looks as though this category can be described concretely as having for its objects triples $(A, T, i: A \otimes \mathbb{Q}/\mathbb{Z} \to T)$ where $A$ is an abelian group, $T$ is a torsion abelian group, and $i$ is an injective homomorphism. Morphisms are given by pairs of homomorphisms $A \to A'$, $T \to T'$ that are compatible with the injective homomorphisms. (Hence, something like a gluing construction.)

The $Ind$-completion of the opposite of finitely presented abelian groups is equivalent to the category of left exact functors $\Phi: \text{Ab}_{fp} \to \text{Set}$, so I'll start there.

Finitely presentable abelian groups are the same as finitely generated $\mathbb{Z}$-modules. Every such module $M$ is a product of a finite power $\mathbb{Z}^n$ and a finite abelian group $F$, so that by left exactness $\Phi(M)$ can be written in the form $\Phi(\mathbb{Z})^n \times \Phi(F)$. It makes sense to consider the restriction of $\Phi$ to the full subcategories $\text{Ab}_{\text{fin}}$ (of finite abelian groups) and $\text{Pow}(\mathbb{Z})$ (of finite powers of $\mathbb{Z}$). Each of these subcategories is finitely complete, and the full inclusions are left exact, so that $\Phi$ restricts to left exact functors on each of these two subcategories.

Interestingly, both of these subcategories happen to be self-dual. Thus in the first case we may as well consider the category of left exact functors $\text{Ab}_{\text{fin}}^{op} \to Set$, that is to say the $Ind$-completion of $\text{Ab}_{fin}$, which is the category of torsion abelian groups. Hence the first restricted functor is given by $\hom(-, T)$, where $T$ is a torsion abelian group.

In the second case, we may as well consider the category of left exact functors $\text{Pow}(\mathbb{Z})^{op} \to Set$. The domain is the Lawvere theory of abelian groups; left exact functors are product-preserving functors, so such a functor must be of the form $\hom(-, A)$ for some abelian group $A$. In fact, in this case left exact functors *coincide* with product-preserving functors, essentially because exact sequences in $\text{Pow}(\mathbb{Z})$ split.

The data $(T, A)$ we have thus extracted do not completely characterize $\Phi$ because we have not taken into account how $\text{Pow}(\mathbb{Z})$ and $\text{Ab}_{\text{fin}}$ interact in $\text{Ab}_{fp}$. We need to consider that $\Phi$ preserves the kernel pair of a morphism $f: \mathbb{Z}^n \to F$ mapping to a finite abelian group. It is easy to convince oneself that $\Phi(f): A^n \to \hom(F, T)$ is a group homomorphism. Since $F$ can be decomposed further as a product of cyclic groups, ultimately we find that the left exactness requirement boils down to the fact that $\Phi$ should preserve exact sequences of the form

$$0 \to \mathbb{Z} \stackrel{- \cdot n}{\to} \mathbb{Z} \to \mathbb{Z}_{n}$$

so that we require exactness of

$$0 \to A \stackrel{- \cdot n}{\to} A \to T_{n}$$

where $T_n$ is the subgroup of elements of $T$ annihilated by $n$. This can be equivalently rephrased as saying that the induced map $A \otimes \mathbb{Z}/(n) \to T \otimes \mathbb{Z}/(n)$ is injective (naturally over all $\mathbb{Z}_n$), and the neatest way of encapsulating all is by passing to the (filtered) colimit over all $\mathbb{Z}/(n)$, which leads to the statement that we have an injective map

$$A \otimes \mathbb{Q}/\mathbb{Z} \to T \otimes \mathbb{Q}/\mathbb{Z} \cong T.$$

There are quite a few details to be checked in this analysis, and I don't claim I've checked every one, but I should imagine this is in the right direction.