Independent vector fields $X,Y$ on $S^3$ with $[X,Y]=Y$

No. If such a pair existed, they would be tangent to a codimension $1$ foliation of $S^3$. Such a foliation must have a Reeb component, in particular, a compact leaf, $T\subset S^3$, which would be a torus (since its tangent bundle would be trivial). If $\xi$ and $\eta$ were the $1$-forms on $T$ dual to the basis $X$ and $Y$ of its tangent bundle, then the above equation would imply that $\mathrm{d}\eta = \eta\wedge\xi$, but this would make a nonvanishing $2$-form on $T$ have vanishing cohomology class in $H^2(T,\mathbb{R})$ (de Rham cohomology), which is clearly impossible.

You can also use this result which appears in the following paper of E. Ghys see corollary I.1.2: Let $G$ be a Lie group homeomorphic to $\mathbb{R}^n$ which acts locally freely on a $n+1$-dimensional manifold $M$. Then the universal cover of $M$ is diffeomorphic to $\mathbb{R}^{n+1}$.

The vector field $X,Y$ define an action of $Aff(\mathbb{R})$ on $S^3$, but the universal cover of $S^3$ is $S^3$ and not $\mathbb{R}^3$.

http://perso.ens-lyon.fr/ghys/articles/actionslocalement.pdf