Wikipedia states :
If an univariate polynomial $p$ has a root (in some field extension) which is also a root of an irreducible polynomial $q$, then $p$ is a multiple of $q$, and thus all roots of $q$ are roots of $p$; this is Abel's irreducibility theorem. This implies that the roots of an irreducible polynomial may not be distinguished through algebraic relations.
The reason the roots of an irreducible polynomial may not be distinguished through algebraic relations ("written out in factored form") is not because of the first sentence above, it is because the polynomial is irreducible in field $F$ therefore has no roots expressible in field $F$.
If $F$ is the algebraic numbers where all polynomial roots are contained and the polynomial is irreducible over $F$ then it has no roots anywhere and is linear.
In an appropriate field, distinguishing the roots is simply a matter of writing out the factors using algebraic relations, with coefficients in that field. So the reason an irreducible polynomial's roots can not be written out in factored form ("distinguished by algebraic relations") is not Abel's irreducibility theorem, it is whether or not the polynomial is irreducible or not in the field in question.
Is this understanding correct or is Wikipedia correct and if so why?
Well, first of all, if you have an irreducible polynomial $f(x)\in F[x]$, then that polynomial is the minimal polynomial over $F$ of any of its roots. If another polynomial in $F[x]$ has a root in common, then by doing the division algorithm, you see that the remainder must be $0$ to not contradict the minimality of $f$, and so $f$ divides the new polynomial.
The reason the roots are "algebraically indistinguishable" is that, given any two roots, there is always an automorphism of the splitting field that sends one root to the other, i.e., the "Galois" group acts transitively on the roots (quotes not needed if the $f$ is separable).