Why are you restricting yourself to $x \in (0,\infty)$ when the domain is for the entire real line? In this case, it seems easier to go with Fourier transforms, i.e.,
$$u(s,t) = \int_{-\infty}^{\infty} dx \, U(x,t) \, e^{i s x}$$
Then the PDE becomes
$$u_t + s^2 u = \sqrt{2 \pi} \, e^{-s^2/2} = e^{-s^2 t}\frac{d}{dt}\left [e^{s^2 t} u \right ]$$
$$u(s,0)=0$$
$$\implies u(s,t) = \sqrt{2 \pi} \, e^{-s^2/2} \frac{1-e^{-s^2 t}}{s^2} $$
Now inverse FT...use the fact that
$$\frac{1-e^{-s^2 t}}{s^2} = \int_0^t dt' \, e^{-s^2 t'}$$
and get
$$U(x,t) = \int_0^t dt' \, (1+2 t')^{-1/2} \, e^{-x^2/[2 (1+2 t')]}$$
Indeed, this problem should be solved with Fourier cosine transform rather than Fourier transform. If you use Fourier transform, you won't be able to take the b.c. at $x=0$ into consideration, and the equation will then be solved in $-\infty<x<\infty$ with implicit b.c.s at $\pm\infty$. (For this part check this post for more information. )
The key point is, when $f(\infty)=0$, Fourier cosine transform
$$
\mathcal{F}_t^{(c)}[f(t)](\omega)=\sqrt{\frac{2}{\pi }}\int _0^{\infty } f(t) \cos (\omega t) d t
$$
has the following property:
$$
\mathcal{F}_t^{(c)}\left[f''(t)\right](\omega)=-\omega^2 \mathcal{F}_t^{(c)}[f(t)](\omega)-\sqrt{\frac{2}{\pi }} f'(0)
$$
You can easily verify this property using integration by parts.
So, by applying this property on your equation, we have
$$
\mathcal{F}_x^{(c)}\left[u^{(1,0)}(t,x)\right](\omega )=k \left(-\omega ^2 \mathcal{F}_x^{(c)}[u(t,x)](\omega )-\sqrt{\frac{2}{\pi }} u^{(0,1)}(t,0)\right)-a k \mathcal{F}_x^{(c)}[u(t,x)](\omega )
$$
Substitute the b.c. into the equation, we obtain a simple initial value problem (IVP) of linear ODE:
$$U'(t)=-a k U(t)-k \omega ^2 U(t)$$
$$U(0)=F$$
where $U(t)=\mathcal{F}_x^{(c)}[u(t,x)](\omega )$, $F=\mathcal{F}_x^{(c)}[f(x)](\omega)$.
If you have difficulty in solving the IVP, check the wikipedia page). Anyway, we can easily find its solution:
$$U(t)=F e^{-k t \left(a + \omega ^2\right)}$$
The last step is to transform back with inverse Fourier Cosine transform
$${\mathcal{F}_\omega^{(c)}}^{-1}[F(\omega)](t)=\sqrt{\frac{2}{\pi }} \int_0^{\infty } F(\omega ) \cos (t \omega) \, d\omega $$
and the solution is
$$
u(t,x)=\sqrt{\frac{2}{\pi }} \int_0^{\infty } e^{-k t \left(a + \omega ^2\right)} \mathcal{F}_x^{(c)}[f(x)](\omega ) \cos (\omega t) \, d\omega
$$
Notice this solution is probably the same as the one given by doraemonpaul. I guess he has just chosen a different convention for Fourier parameters.
Best Answer
If $f : \mathbb R\to\mathbb C$ is a "nice" function (i.e. $f\in H^1$ or $f\in C^1$ with $f(\pm\infty) = 0$) the Fourier transform of its derivative is by partial integration $$ \widehat{f'}(\omega) = \int f'(x)e^{-2\pi ix\omega}\,dx = 2\pi i\omega\int f(x)e^{-2\pi ix\omega}\,dx = 2\pi i\omega\cdot\hat f(\omega). $$ So if $U(\omega,t)$ denotes the Fourier transform of $u(x,t)$ with respect to $x$, applying the Fourier transform with respect to $x$ to $u_t = u_{xx} + \alpha u_x$ gives $$ U_t = -4\pi^2\omega^2 U + 2\pi i\omega\alpha U = (-2\pi + i\alpha\omega)2\pi U. $$