What the question means, I think, is to take the Fourier Transform in the spatial variable and the Laplace Transform in the time variable.
First, taking the Laplace Transform $\mathcal{L}[u(x,t](s) = w(x, s)$ and letting $u_0(x) := u(x, 0)$ be a known initial condition, we obtain:
$$sw(x,s) - u_0(x) -\kappa\frac{\partial^2 w(x,t)}{\partial x^2} = S_0 \delta(x)$$
which is equivalent to
$$\left(s - \kappa\frac{\partial^2}{\partial x^2}\right)w(x,s) = S_0\delta(x) + u_0(x)$$
Taking the Fourier Transform of this equation, letting $\mathcal{F}[v(x, s)](k) = v(k, s)$, we obtain
$$\left(s+\kappa k^2\right)v(k, s) = S_0 + \int_{-\infty}^{\infty}u_0(x)e^{-ikx}dx$$
or, equivalently
$$v(k,s) = \frac{1}{s + \kappa k^2}\left(S_0 + \int_{-\infty}^{\infty}u_0(x)e^{-ikx}dx\right)$$
(depending on your Fourier Transform convention, various factors of ${2\pi}^{-1}$ may appear).
Now, we only have to invert the two transforms, and then we are done.
The spatial stuff looks pretty messy, but the Laplace Transform can be done easily, as we only have one $s$ in there.
In fact, we have a function that looks something like this:
$$v(k, s) = \frac{F(k)}{s + a(k)}$$
which has an Inverse Laplace Transform of $p(k, t) = \mathcal{L}^{-1}[v(k,s)](t) = F(k)\exp(-a(k)t)$, so we are left with
$$p(k, t) = \exp\left(-\kappa k^2 t\right)\left(S_0 + \int_{-\infty}^{\infty}u_0(x)e^{-ikx}dx\right)$$
The Inverse Fourier Transform of the first term can be reduced to calculating a Gaussian integral, giving us the typical Gaussian solution one would expect from the Heat Equation as a homogeneous solution. This calculation is outlined below:
\begin{align}
u_H(x, t) &= \mathcal{F}[S_0 \exp\left(-\kappa k^2 t\right)](x, t)\\
&= \frac{S_0}{2\pi}\int_{-\infty}^\infty \exp(-\kappa t k^2 + i x k)dk\\
&= \frac{S_0}{2\pi}\int_{-\infty}^\infty \exp\left(-\kappa t\left(k^2 - \frac{ixk}{\kappa t}\right)\right)dk\\
&= \frac{S_0}{2\pi}\int_{-\infty}^\infty\exp\left(-\kappa t\left(k-\frac{ix}{2\kappa t}\right)^2 - \frac{x^2}{4\kappa t}\right)dk\\
&= \frac{S_0}{2\pi}\exp\left(-\frac{x^2}{4\kappa t}\right)\int_{-\infty}^\infty
\exp\left(-\kappa t\left(k-\frac{ix}{2\kappa t}\right)^2\right)dk
\end{align}
Letting $z := k - \frac{ix}{2\kappa t}$, we get that $dk = dz$. The contour on integration shifts to a line parallel to the real axis, but it is not too difficult to show that if one were to consider a a rectangular closed contour the runs along both this line and the real axis, the contribution of the short edges of the rectangle would go to zero as the length of the line approaches infinity. Thus, we can shift the integral back onto the real axis, as the integral along the real axis and along the line parallel to it must be equal in value.
\begin{align}
u_H(x, t) &= \frac{S_0}{2\pi}\exp\left(-\frac{x^2}{4\kappa t}\right)\int_{-\infty}^\infty
\exp\left(-\kappa tz^2\right)dz\\
&= \frac{S_0}{2\pi}\exp\left(-\frac{x^2}{4\kappa t}\right)\sqrt{\frac{\pi}{\kappa t}}\\
&= \frac{S_0}{\sqrt{4\pi\kappa t}}\exp\left(-\frac{x^2}{4\kappa t}\right)
\end{align}
This is the known solution to the Heat Equation as a homogeneous solution.
The inhomogeneous solution can only be written down in integral form, as we do not know anything about the initial distribution $u_0(x)$. It is given by:
\begin{align}
u_I(x, t) &= \frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^{\infty}\exp\left(-\kappa k^2 t\right)e^{ikx} e^{-iky} u_0(y) dy dk\\
&= \frac{1}{2\pi}\int_{-\infty}^\infty u_0(y)\int_{-\infty}^\infty \exp\left(-\kappa t k^2 + ik(x-y)\right)dkdy\\
&= \frac{1}{\sqrt{4\pi \kappa t}}\int_{-\infty}^\infty u_0(y)\exp\left(-\frac{(x-y)^2}{4\kappa t}\right) dy
\end{align}
which is preciously the convolution of the initial condition $u_0(x)$ with the fundamental homogeneous solution $u_H(x, t)$.
The final solution is then given as the sum of the homogeneous and the inhomogeneous solution.
EDIT: The above was done for a general $u_0(x)$, because I missed that $u_0(x)$ had been specified in the questions. Given that we have $u_0(x) = \delta(x)$, we can calculate the final integral easily:
\begin{align}
u_I(x,t) &= \frac{1}{\sqrt{4\pi \kappa t}}\int_{-\infty}^\infty \delta(y)\exp\left(-\frac{(x-y)^2}{4\kappa t}\right) dy\\
&= \frac{1}{\sqrt{4\pi \kappa t}}\exp\left(-\frac{x^2}{4\kappa t}\right)
\end{align}
This gives us the full solution
$$u(x,t) = \frac{S_0 + 1}{\sqrt{4\pi \kappa t}}\exp\left(-\frac{x^2}{4\kappa t}\right)$$
Best Answer
$\displaystyle \int_0^\infty f(t) e^{-st} dt$ should be thought as the Laplace transform of a function defined only on $t \in [0;\infty[$ or as the bilateral Laplace transform of a function $f(t)$ being $0$ for $t<0$
the Laplace transform $$\int_a^\infty f(t) e^{-st} dt$$ is injective (in the almost everywhere sense) on functions defined on $[a;\infty[$ for which it converges when $Re(s)\to \infty$
(often $a=0$ but depending on the function it can be another value, and it can even be $-\infty$ but in that case it becomes the bilateral Laplace transform)
the bilateral Laplace transform is injective on functions defined on $\mathbb{R}$ for which it converges when $Re(s) = c$ (for a fixed $c$) and there it coincides with the Fourier transform of $f(t) e^{-ct}$ .
note that $-\int_{-\infty}^0 e^{-st} dt = 1/s$ for $Re(s) < 0$ and $\int_{0}^\infty e^{-st} dt = 1/s$ for $Re(s) > 0$ :
$\implies$ many functions have different inverse Laplace transforms, depending on the assumed domain of convergence.
so it is not injective anymore without the convergence assumption, this is why the domain of convergence is so important.