I'd say the most intuitive way is to integrate the electrostatic equations (written in Heaviside-Lorentz units and for the vacuum),
\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho.
The first equation tells us that the electric field has a scalar potential,
\vec{E}=-\vec{\nabla} \cdot \Phi.
Gauß's Law then becomes the Poisson Equation,
\Delta \Phi=-\rho.
In your case of a homogeneously charged spherical shell you have
\rho(\vec{x})=\frac{Q}{4 \pi R^2} \delta(r-R),
where r=|\vec{x}|.
The charge distribution is spherically symmetric and thus one can expect that the potential is a function of r only. Writing the Laplace operator in terms of spherical coordinates our Poisson equation reads
\frac{1}{r} \frac{\mathrm{d}^2}{\mathrm{d} r^2} (r \Phi)=-\frac{Q}{4 \pi R^2} \delta(r-R). \qquad (*)
For r \neq R the right-hand side is 0 and the general solution thus reads
with integration constants A and B, which are to be chosen different for r<R and r>R due to the singularity of the \delta distribution. We have to find the constants from appropriate boundary conditions.
Start with r<R. Since there are no sources or other singularities there, we must have A=0 in that region, because otherwise the potential would have a singularity in the origin. Thus we have
\Phi(r)=B \quad \text{for} \quad r<R.
For r>R we only demand that the potential vanishes at infinity. This fixes simply the overall additive constant of the potential, which has no physical meaning anyway.
So our solution reads
\Phi(r)=B \Theta(R-r)+\frac{A}{r} \Theta(r-R).
The potential should be continuous at r=R since only its second derivative should have a \delta-distribution like singularity. The first derivative must thus have a jump. So we must have B=A/R. And thus
\Phi(r)=A \left [\frac{1}{R} \Theta(R-r)+\frac{1}{r} \Theta(r-R) \right ].
To determine finally A, we have to multiply the equation (*) by r and integrate over an infinitesimal interval around r=R. This gives
\left ([r \Phi(r)]' \right )_{r=R+0^+}-\left ([r \Phi(r)]' \right )_{r=R-0^+}=-\frac{Q}{4 \pi R}.
Evaluating the left-hand side gives
-A/R=-\frac{Q}{4 \pi R} \; \Rightarrow \; A=\frac{Q}{4 \pi}.
The potential thus reads
\Phi(r)=\frac{Q}{4 \pi} \left [\frac{\Theta(R-r)}{R}+\frac{\Theta(r-R)}{r} \right ].
Inside the spherical shell the potential is constant and thus the field vanishes there. Outside you have a Coulomb potential as if the charge is sitting at the center.
Of course you can also evaluate the potential using the Green's-function method, according to which you just integrate over all points of your charge distribution, using Coulomb's Law, i.e.,
\Phi(\vec{x})=\frac{1}{4 \pi} \int_{V} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.
In your case the integral simplifies to an integral along the spherical surface,
\Phi(\vec{x})=\frac{1}{4 \pi} \frac{Q}{4 \pi R^2} R^2 \int_0^{2 \pi} \mathrm{d} \varphi' \int_0^{\pi} \mathrm{d} \vartheta' \frac{\sin \vartheta'}{|R \vec{e}_r(\vartheta',\varphi')-\vec{x}|}.
Here I've used spherical coordinates (\vartheta',\varphi') to parametrize the spherical shell and set the radial unit vector to \vec{e}_r(\vartheta',var\phi')=(\cos \varphi' \sin \vartheta',\cos \varphi' \sin \vartheta',\cos \vartheta'). Now, you can choose the spherical coordinates such that \vec{x}=r \vec{e}_z, i.e., that this vector points along the polar axis of the spherical coordinates. Then you have
|R\vec{e}_r-\vec{x}|^2=(R^2 + r^2 - 2 R r \cos \vartheta').
The integral over \varphi' thus just gives a factor 2 \pi, and you get
\Phi(\vec{x})=\Phi(r)=\frac{Q}{8 \pi} \int_0^{\pi} \mathrm{d} \vartheta' \frac{\sin \vartheta'}{\sqrt{R^2+r^2-2 R r \cos \vartheta'}}.
Now you substitute u=\cos \vartheta', \quad \mathrm{d} u=-\mathrm{d} \vartheta' \sin \vartheta', which gives
\Phi(r)=\frac{Q}{8 \pi} \int_{-1}^{1} \mathrm{d} \vartheta' \frac{1}{\sqrt{R^2+r^2-2 R r u}}.
This you can integrate in closed form, giving
\Phi(r)=\frac{Q}{8 \pi} \frac{R+r-|R-r|}{r R} = \frac{Q}{4 \pi} \left [\frac{\Theta(R-r)}{R}+ \frac{\Theta(r-R)}{r} \right].
This is of course the same result as with the direct calculation of the Poisson equation.
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