With such simple initial data, the most helpful way is to find the solution using the Poisson formula
\begin{align}
u(x,t)=\frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}ae^{-by^2}e^{-\frac{(x-y)^2}{4t}}dy=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}e^{-(b+\frac{1}{4t})\bigl(y-\frac{x}{1+4bt}\bigr)^2}dy\\
=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}e^{-(b+\frac{1}{4t})y^2}dy=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{\pi(1+4bt)}}\int\limits_{-\infty}^{\infty}e^{-z^2}dz=\frac{ae^{-\frac{bx^2}{1+4bt}}}{\sqrt{1+4bt}}
\end{align}
which readily implies that
$$
\lim_{|x| \to \infty} u_x(x,t) = 0 = \lim_{|x| \to \infty} u(x,t).\tag{$\ast$}
$$
But the fundamental fact is that $(\ast)$ holds with any bounded initial data $u_0(x)\overset{\rm def}{=}u(x,0)$ vanishing at infinity, i.e.,
$$
\sup_{\mathbb{R}}|u_0|=M<\infty,\qquad\lim_{|x| \to \infty} u_0(x) = 0.
$$
Indeed, $\forall\,\varepsilon>0\;\exists\,R_{\varepsilon}>0\,\colon\; |u_0(x)|<\varepsilon/2\;\forall\,x\in\mathbb{R}$ with modulus $ |x|>R_{\varepsilon}\,.$ Hence
\begin{align}
|u(x,t)|\leqslant\frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}|u_0(y)|e^{-\frac{(x-y)^2}{4t}}dy=\frac{1}{\sqrt{4\pi t}}\biggl(\int\limits_{|y|<R_{\varepsilon}}+\int\limits_{|y|>R_{\varepsilon}}\biggr)\\
\leqslant\frac{M}{\sqrt{4\pi t}}\int\limits_{-R_{\varepsilon}}^{R_{\varepsilon}}e^{-\frac{(x-y)^2}{4t}}dy+\frac{\varepsilon/2}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}e^{-\frac{(x-y)^2}{4t}}dy\leqslant \frac{M}{\sqrt{4\pi\delta}}\int\limits_{-R_{\varepsilon}}^{R_{\varepsilon}}e^{-\frac{(x-y)^2}{4T}}dy+\varepsilon/2\tag{$\ast\ast$}
\end{align}
for all $x\in\mathbb{R}$ and $t\in [\delta,T]$, with arbitrary fixed positive $\delta<T$.
Since the last integral in $(\ast\ast)$ is vanishing as $|x|\to\infty$, there is some $d>0$ such that
$$
\frac{M}{\sqrt{4\pi\delta}}\int\limits_{-R_{\varepsilon}}^{R_{\varepsilon}}e^{-\frac{(x-y)^2}{4T}}dy<\varepsilon/2\quad\forall\,x\colon\;|x|>d
$$
which implies that $|u(x,t)|<\varepsilon/2+\varepsilon/2=\varepsilon$ when $|x|>d$ and $t\in [\delta,T]$. Thus it has been established that
$$
\lim_{|x| \to \infty} u(x,t)=0\quad\forall\, t\in (0,\infty).
$$
The rest can be done in a similar manner, since
$$
u_x(x,t)=-\frac{1}{2t\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}(x-y)u_0(y)e^{-\frac{(x-y)^2}{4t}}dy,
$$
while
$$
\frac{1}{2t\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}|x-y|e^{-\frac{(x-y)^2}{4t}}dy=
\frac{1}{2t\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}|y|e^{-\frac{y^2}{4t}}dy=
\frac{1}{t\sqrt{4\pi t}}\int\limits_{0}^{\infty}ye^{-\frac{y^2}{4t}}dy=
\frac{1}{\sqrt{\pi t}}.
$$
If
$u(x,t) = e^{\alpha x−\beta t}v(x,t), \tag 1$
then
$u_t(x, t) = -\beta e^{\alpha x−\beta t}v(x,t) + e^{\alpha x−\beta t}v_t(x,t), \tag 2$
$u_x(x, t) = \alpha e^{\alpha x−\beta t}v(x,t) + e^{\alpha x−\beta t}v_x(x,t), \tag 3$
$u_{xx}(x, t) = \alpha^2 e^{\alpha x−\beta t}v(x,t) + \alpha e^{\alpha x−\beta t}v_x(x,t) + \alpha e^{\alpha x−\beta t}v_x(x,t) + e^{\alpha x−\beta t}v_{xx}(x,t)$
$= \alpha^2 e^{\alpha x−\beta t}v(x,t) + 2\alpha e^{\alpha x−\beta t}v_x(x,t) + e^{\alpha x−\beta t}v_{xx}(x,t); \tag 4$
we are given that $u(x, t)$ satisfies
$u_t=Du_{xx}+Au_x+Bu,\quad x\in\Bbb R, t>0; \tag 5$
we substitute in the formulas (1)-(4):
$-\beta e^{\alpha x−\beta t}v + e^{\alpha x−\beta t}v_t = D( \alpha^2 e^{\alpha x−\beta t}v + 2\alpha e^{\alpha x−\beta t}v_x + e^{\alpha x−\beta t}v_{xx})$
$+ A(\alpha e^{\alpha x−\beta t}v + e^{\alpha x−\beta t}v_x) + Be^{\alpha x−\beta t}v, \tag 6$
and cancel out
$e^{\alpha x−\beta t} \ne 0: \tag 7$
$-\beta v + v_t = D( \alpha^2 v + 2\alpha v_x + v_{xx}) + A(\alpha v + v_x) + Bv, \tag 8$
and gather $v$ and its derivatives into like terms:
$v_t = Dv_{xx} + (2D \alpha + A) v_x + (D\alpha^2 + A\alpha + (B + \beta))v, \tag 9$
which is reducible to a heat equation of the form
$v_t = D v_{xx} \tag{10}$
provided we may take $\alpha$ and $\beta$ satisfying
$2D\alpha + A = 0, \tag{11}$
$D\alpha^2 + A\alpha + B + \beta = 0; \tag{12}$
these two equations are easy to solve; (11) yields
$\alpha = -\dfrac{A}{2D}, \tag{13}$
and when this is substituted into (12):
$D \left ( -\dfrac{A}{2D} \right)^2 + A \left ( -\dfrac{A}{2D} \right ) + B + \beta = 0, \tag{14}$
or
$\dfrac{A^2}{4D} - \dfrac{A^2}{2D} + B + \beta = 0, \tag{15}$
yielding
$\beta = \dfrac{A^2}{4D} - B; \tag{16}$
these values of $\alpha$ (13) and $\beta$ (16) result in the heat equation (10) when inserted into (9).
Best Answer
Since $e^{\alpha t}$ is never zero, multiplying by $e^{\alpha t}$ is an invertible operation. That is, for every function $f(t, x)$ and every $\alpha$, there exists a unique function $\omega(t, x, \alpha)$ such that $f(t, x) = e^{\alpha t}\omega(t, x, \alpha)$. Since every function has this decomposition, then in particular our solution $u(t, x)$ also has that decomposition.
You might have noticed I wrote $\omega(t, x,\alpha)$ instead of $\omega(t,x)$. Indeed, $\omega$ will be a different function for different $\alpha$. However, because $\alpha$ is just a constant, it's conventional to leave out the $\alpha$ from the argument of $\omega$. Just like it's conventional to write $u(t, x)$ instead of the technically more correct $u(t, x, \gamma)$.