$E (\mathbb Q^{ab})/\operatorname{tors}$ is a sum of countably many copies of $\mathbb Z$. Iterating, we form a $\mathbb Z$-basis of the whole group.
To prove this, take a countable basis $x_1,x_2,\dots$ of $E (\mathbb Q^{ab})/\operatorname{tors}\otimes \mathbb Q$. Suppose we show that $\oplus_{i=1}^n x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$ is finitely generated. Then it is necessarily isomorphic to $\mathbb Z^n$. Moreover, the only embeddings $\mathbb Z^n \to \mathbb Z^{n+1}$ where the elements of $\mathbb Z^n$ don’t get any more divisible are isomorphic to the standard embedding. So if $x_1,\dots,x_n$ form a $\mathbb Z$-basis of $\oplus_{i=1}^n x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$, we can change $x_{n+1}$ so it generates the same $\mathbb Q$-subspace and also $x_1,\dots,x_{n+1}$ form a $\mathbb Z$-basis of $\oplus_{i=1}^{n+1} x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$.
Now let’s check the claim. Let $k$ be a cyclotomic field containing $x_1,\dots,x_n$ and all the torsion points of $E (\mathbb Q^{ab})/\operatorname{tors}$. Let $N$ be the maximum order of a torsion point. Let $y$ be any point of $E (\mathbb Q^{ab})/\operatorname{tors}$ such that $m y \in E(k)$ for some natural number $m$. We will check that $N y \in E(k)$. It will follow that$ \oplus_{i=1}^n x_i \mathbb Q \cap E (\mathbb Q^{ab})/\operatorname{tors}$ is contained in $(1/N) E(k)$, which is finitely generated, so is finitely generated.
To check this, let $y’$ be any Galois conjugate of $y$ relative to $k$. Because $\mathbb Q^{ab}/\mathbb Q$ is Galois, $y’ \in E(\mathbb Q^{ab})$, so $y’-y \in E(\mathbb Q^{ab})$. Because the Galois conjugation is relative to $k$, $m y’=my $, so $m(y’-y) =0$ and $y’-y$ is torsion, hence it is $N$-torsion, so $Ny’ =Ny$. Because this is true for all $y’$, $Ny$ is Galois-invariant and thus is in $E(k)$.