This is indeed correct and can be made rigorous, assuming that the integral converges sufficiently well for all $u_0$, which in turn depends on the boundary conditions that are imposed for the Laplacian.
Assume that $\int_0^\infty \Vert u(\cdot,t) \Vert dt < \infty$ for all $u_0$, for a suitable norm (e.g. the $L^2$ norm). By a theorem of Datko and Pazy, this implies that the spectrum of $\Delta$ is contained in the left half plane and bounded away from the imaginary axis. Now write formally $A = \Delta$ and $u(\cdot,t) = e^{At}u_0$. You are then computing
$$
\int_0^\infty e^{At} u_0 dt = (-A)^{-1} u_0 = (-\Delta)^{-1} u_0 \, .
$$
More generally, for $\lambda$ in a suitable right half plane,
$$
\int_0^\infty e^{At} e^{-\lambda t} dt = (\lambda I - A )^{-1}
$$
that is, Laplace transforms of the operator semigroup $\left( e^{At} \right)_{t \ge 0}$ are resolvents of the generator $A$ of the semigroup.
All this can be made rigorous using semigroup theory.
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
The heat equation (also known as diffusion equation) conserves total mass, which by definition is the integral $M (t) = \int_{-\infty}^\infty u(x,t)\,dx$. (This can be proved by taking time derivative of $M$ and using the PDE.) Since $\delta$ is a unit mass, in order to satisfy $u(x,0)=\delta(x)$ we need $M$ to be $1$.
This is why we normalize the fundamental solution $$u(x,t)=\frac{c}{\sqrt{t}}e^{-x^2/(4t)}$$ so that its mass is $1$. Which leads to $$c^{-1} = \int_{-\infty}^{\infty} e^{-x^2/4}\,dt =\sqrt{4\pi}$$ Above, the mass is computed at time $t=1$ and equated to $1$. It could be computed in any other moment $t_0>0$; the result is the same.