If you take $w\in C^\infty([0,T])$, then the endpoint map depends smoothly on $w, a, b$. This is most easily seen using convenient calculus: am mapping is smooth if it maps smooth curves to smooth curves, and the space of smooth curves in $C^\infty([0,T])$ is just $C^\infty(\mathbb R\times [0,T])$. So the result follows from smooth depended of solutions of ODE on one further parameter. See this Wikipedia page and literature cited there.
For $L^2([0T])$ one has to work harder to smoothness.