Turing-recognizable languages closed under star operation
You're confusing the definition of $A^{\ast}$ and $A$. It is only $A$ in your example that contains strings of $a$s whose lengths are powers of two. $A^{\ast}$ does not have this restriction - it has that restriction that its strings are made up of strings from $A$. 

So the individual components of each string in $A^{\ast}$ have to have lengths of powers of two (because they're from $A$), but together we only get that for every string $a^{m}$ in $A^{\ast}$ there is some set $N = \{n_{1},\ldots,n_{k}\}$ (different for each string) such that $m = \sum_{n_{i} \in N} 2^{n_{i}}$.