The answer seems to me Yes. There exists $u\in C^{1,3}([0, T]\times \mathbb R)$.
For simplicity, we set $g = 0$. Let's use the norm $$|u|_{1,3} = |u|_0 + |\partial_x u|_0 + |\partial_{xx} u|_0 + |\partial_t u|_0.$$ For $\lambda \in [0,1]$, we define $$L_\lambda u = \partial_t u - \partial_{xx} u - \lambda (\partial_x u - cu).$$ Then, the following a priori estimates hold, by setting $L_\lambda u = f \in C^{0,2}([0, T]\times \mathbb R)$ $$|u|_0 \le K(|c|_0) |f|_0.$$ $$|\partial_x u|_0 \le K(|c|_0, |\partial_x c|_0)(|\partial_x f|_0 + |f|_0).$$ $$|\partial_{xx} u|_0 \le K(|c|_0, |\partial_x c|_0, |\partial_{xx} c|_0)(|\partial_{xx} f|_0 + |\partial_x f|_0 + |f|_0).$$ $$|\partial_{t} u|_0 \le K(|c|_0, |\partial_x c|_0, |\partial_{xx} c|_0)(|\partial_{xx} f|_0 + |\partial_x f|_0 + |f|_0).$$ Therefore, we get $$|u|_{1,3} \le K(|c|_0, |\partial_x c|_0, |\partial_{xx} c|_0) |L_\lambda u|_{0,2}.$$ Since $L_0: C^{1,3} \mapsto C^{0,2}$ is onto, the method of continuity yields the solvability in $C^{1,3}$.