The first attempt with method of continuity does not go through. This is second attempt based on the hints of other replies. The answer seems to me Yes. There exists $u\in C^{1,2}([0, T]\times \mathbb R)$.
Let $$F(c, u) = F[c] u = \partial_t u - \partial_{xx} u - \partial_x u + cu.$$ We use the following claims. (C1) and (C2) are standard, and (C3) will be proved later.
(C1) The operator $F: C^{0,0} \times C^{1,2} \mapsto C^{0,0}.$ is a continuous mapping
(C2) If $c, f$ are Holder continuous, then exists unique $u\in C^{1,2}$, s.t. $F(c, u) = f$, i.e. $u = F^{-1}[c] f$ is well defined.
(C3)For $c_i, f_i \in C^{1,3}$ satisfying $|c_i|_{0,2} + |f_i|_{0,2} < K$ with $i = 1, 2$, we have
$$|F^{-1}[c_1] f_1 - F^{-1} [c_2] f_2 |_{1,2} \le\Psi(K, T)( |f_1 - f_2|_{0,2} + |c_1 - c_2|_{0,2})$$ for some strictly increasing continuous function $\Psi(K, T)$ with $\Psi(0, T) = \Psi(K, 0) = 0.$
Now, let $c, f\in C^{0,2}$ be given. Then, since $C^{1,3}$ is dense in $C^{0,2}$, there exists $c_n, f_n \in C^{1,2}$, s.t. $$|c_n -c|_{0,2} + |f_n -f|_{0,2} \to 0$$ and $$|c_n|_0 + |f_n|_0 \le 2(|c|_0 + |f|_0) := K.$$ We denote $u_n = F^{-1} [c_n] f_n$. Then, $u_n$ is Cauchy in $C^{1,2}$ since by (C3) $$|u_n - u_m|_{1,2} \le \Psi(K, T) (|f_n - f_m|_{0,2} + |c_n - c_m|_{0,2}).$$ So there exists $u\in C^{1,2}$ s.t. $|u_n - u|_{1,2} \to 0$. Now we can verify $u$ is the solution by checking $$F(c, u) = \lim_n F (c_n, u_n) = \lim_n f_n = f.$$ In the above, we used (C1).
The remaining part is the proof of (C3). For $c_i, f_i \in C^{1,3}$ satisfying $|c_i|_{0,2} + |f_i|_{0,2}< K$ with $i = 1, 2$, $u_n = F^{-1} [c_n] f_n$ for $n=1 , 2$ is a classical solution by (C2) and $v_n (t, x) = u_n(T - t, x)$ has probability representation of the form $$v_n(t, x) =\mathbb E \Big[ \int_t^{T} \exp\{- \int_t^{s} c_n(r, X^{t,x}(r)) dr\} f_n(s, X^{t,x}(s) )ds\Big] $$ where $$X^{t, x} (s)= x + (t-s) + W(s) -W(t).$$ By direct approximation, one can have $$|v_1 - v_2|_0 \le KT^2 e^{KT}(|f_1 - f_2|_0 + |c_1 - c_2|_0) .$$ This also holds for \begin{equation} \label{eq:01} |u_1 - u_2|_0 \le KT^2 e^{KT}(|f_1 - f_2|_0 + |c_1 - c_2|_0) := \Psi(K, T)(|f_1 - f_2|_0 + |c_1 - c_2|_0) . \end{equation} Next, we can check that, by (C2) $\bar u_n = \partial_x u_n$ is the classical solution of $$\partial_t \bar u_n = \partial_x \bar u_n + \partial_{xx} \bar u_n - c_n\bar u_n - \partial_x c_n \cdot u_n + \partial_x f_n$$ with $\bar u_n(0, x) = 0.$ Similarly, we have $$|\partial_x (u_1 - u_2)|_0 \le \Psi(K, T)(|(- u_1\partial_x c_1 + \partial_x f_1) - (- u_2\partial_x c_2 + \partial_x f_2)|_0 + |c_1 - c_2|_0).$$ Combined with the earlier estimation on $|u_1 - u_2|_0$, we have $$|u_1 - u_2|_{0, 1} \le \Psi(K, T) (|f_1 - f_2|_{0, 1} + |c_1 - c_2|_{0,1}).$$ Using exactly the same approach, we have $$|u_1 - u_2|_{0, 2} \le \Psi(K, T) (|f_1 - f_2|_{0, 2} + |c_1 - c_2|_{0,2}).$$ Together with original equation $\partial_t u = ...$, we have final estimation $$|u_1 - u_2|_{1, 2} \le \Psi(K, T) (|f_1 - f_2|_{0, 2} + |c_1 - c_2|_{0,2}).$$ This completes the proof of (C1).