I claim for every nontorsion $x \in E(\mathbb Q^{ab})$, there is some $n$ such that $x$ is not $m$-divisible for any $m>n$.

Indeed, let $k$ be a cyclotomic field over which $E$ and all the torsion of $E(\mathbb Q^{ab})$ are defined. Without loss of generality, $x$ is indivisible in $E(k)$. Let $n$ be the maximal order of a torsion point of $E(\mathbb Q^{ab})$.

If $y$ is an $m$th root of $x$, then for each prime $p$ dividing $m$, $y^{m/p}$ is not defined over $k$, so $y$ has some Galois conjugate $y'$ with $y^{m/p} \neq y'^{m/p}$ but $y^m = y'^m$. Then if $y \in E(\mathbb Q^{ab})$, $y'$ is as well because it's Galois, so $(y/y')$ is a torsion point or order dividing $\gcd(m,n)$. If $m$ is not a multiple of $n$, then $\gcd(m,n)$ is a proper divisor of $m$. Taking $p$ dividing $m/\gcd(m,n)$, we see that $(y/y')^{m/p}$ is trivial, contradiction.

So in fact the group is closer to a sum of copies of $\mathbb Z$. But it is not clear from this sort of "local" argument whether it is a sum of copies of $\mathbb Z$.