alt Hacker Newshere's another version directly from the horse's mouth : "Consider the natural map π: P¹ × Sym²(P¹) → Sym³(P¹), (p, {q,r}) ↦ {p,q,r}. Let R be its ramification divisor and let H ⊂ Sym³(P¹) ≅ P³ be a hyperplane tangent but not osculating to the small diagonal; identify X := (P¹ × Sym²(P¹)) \ (R ∪ π⁻¹(H)) ≅ A³ and Y := Sym³(P¹) \ H ≅ A³. Take π|X: X → Y." This is in fact so simple if correct that someone should have found it after all...
My Claude found a similar description (it phrased it in terms of the natural map from "cubics with a choice of root" to "cubics"). The part that seems not at all simple or obvious is the fact that X is isomorphic to A^3. In your presentation (and more or less similarly in the one my Claude found), X is given as P1 x P2 minus a reducible hypersurface, also I think R itself is reducible since it contains points of the form (p, {p, q}) and (p, {q, q}). Then it takes some calculation to identify X with A^3.