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 if the state domain is changed to 1-torus $\mathbb T$ from $\mathbb R$.

- [Conclsion] There exists $u\in C^{1,2}([0, T]\times \mathbb T)$.

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 $\Psi(0, T) = \Psi(K, 0) = 0.$

(C4) if $\phi \in C^{0,2}$, then there exists $\phi_n \in C^{1,3}$ s.t.
$|\phi_n - \phi|_{0,2} \to 0$. This is the place why the domain is changed to $\mathbb T$. Indeed, one can use polynomial $p_n$ approximate $\partial_{xx} \phi$ by stone-weirstauss and take integral twice to get $\phi_n$.

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).