Strictly speaking, Young's modulus is not always greater than the shear modulus, but it does tend to work out that way. You can see the reason why if you look at the relation between the two quantities (and Poisson's ratio).
$$
G = \frac{E}{2(1+\nu)}
$$
Combined with the knowledge that $\nu$ can be anywhere in the range $(-1, \frac{1}{2})$, one can see that G can be greater than E for $\nu < -1/2$. That being said, materials with such a negative Poisson's ratio are extremely uncommon, and it is safe to assume that the shear modulus is less than half of Young's modulus.
For an explanation of why the Poisson's ratio must fall within the above range, I invite you to check out some of my previous answers.
High-level: Range of poissons ratio
Detailed explanation: Limits of Poisson's ratio in isotropic solid
For now, let's ignore the expansion of the container due to the heating and just focus on the stress in the wall of the pressure vessel. I will also examine the case of a thin-walled, spherical vessel, but the same procedure may be applied for other geometries.
Compute Stress State
First, you must compute the pressure in the walls. To do this, imagine a cutting plane through any diameter of the vessel. Now, balance forces on one half of the vessel. In one direction, you have the force of the fluid pressure, which totals to $F = \pi R^2 P$. In the other direction, the only balancing force is the stress in the walls acting over the exposed cross-section. The force here equals $F = 2\pi R t \sigma_{wall}$, where $t$ is the thickness of the wall. By setting these forces equal, we can compute that the stress in the wall is
$$
\sigma_{wall} = \frac{P R}{2t}.
$$
Now, due to the symmetry of the problem, and making the (valid) simplification that of zero through-thickness stress, we can write the full stress state of any point in the wall:
$$
\mathbf{\sigma} = \sigma_{wall}
\left(
\mathbf{e}_{\theta}
\otimes
\mathbf{e}_{\theta}
+
\mathbf{e}_{\phi}
\otimes
\mathbf{e}_{\phi}
\right)
+
0\;
(\mathbf{e}_{r}
\otimes
\mathbf{e}_{r})
$$
Check Yield Criterion
To compute the point where yield happens, you must compute the Mises equivalent tensile stress. There are a variety of equivalent definitions, but all will lead you to the conclusion that in order to prevent plastic deformation, $\sigma_{wall} < \sigma_y$, where $\sigma_y$ is the material's uniaxial tensile yield stress. Thus, the maximum pressure the vessel can contain before yield is
$$P < \frac{2\,t\,\sigma_{y}}{R}$$
Other Considerations
You still have to consider fracture, but that's a separate discussion that is more suited to an engineering class. (If this is what you're looking for, I can go there...) With this comes the "leak before break" criterion, which puts an upper bound (counter-intuitive, but it checks out) on the safe thickness of the vessel walls.
Another point to keep in mind is that the pressure in the tank will change with temperature. Be sure to account for this in your analysis with $PV = nRT$ or some other appropriate state equation.
As others have mentioned in the comments on your question, there are very strict codes that put factors of safety on essentially every aspect of the tank design. This would be the place to start if you're actually going to build something.
Best Answer
The modulus of elasticity (Young's Modulus), $E$, and the shear modulus, $G$, are related by the equation:
$$G=\frac{E}{2(1+ν)}$$
Where $ν$ is Poisson's ration = -(lateral strain)/longitudinal strain).
As you can see, the two are proportional to one another. I personally never heard of the shear modulus being called modulus of rigidity and I agree with you it doesn't seem to make sense to call one "rigidity" and the other "elasticity" when the are linearly related. You'd think they would both be called rigidity or elasticity, but not the opposite.
Young's modulus involves longitudinal stress/strain (tension/compression). The shear modulus involves transverse or lateral stress/strain (shear), so it is logical they are related to each other by Poisson's ratio (ratio of lateral to longitudinal strain). You can also see this because when you longitudinally compress or stretch something it laterally expands and contracts, respectively, as well.
The only reason I can think of is to avoid confusion in the use of terms. If both the shear modulus and Young's modulus were referred to as "modulus of elasticity", or, for that matter, "modulus of rigidity" how would we know which modulus was being referred to? What I was trying to say is there should be no technical reason for the difference in terms for $G$ and $E$, since they both refer to resistance to deformation (lateral and longitudinal).
Hope this helps.