Solution

Let us write the boundary condition for the normal components of the electric displacement field:
$$D_{2n} - D_{1n} = \sigma$$
where $\sigma$ is the free surface charge density at the boundary.
Since $D = \varepsilon_0 E$, for the first case ($E_1 = E$, $E_2 = 2E$) we have:
$$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (2E) - \varepsilon_0 E = \varepsilon_0 E$$

To find the pressure, consider a thin cylindrical shell at the boundary of the two media. Let us isolate a small portion of the surface with charge $\Delta q = \sigma \Delta S$. Let the field generated by this specific portion be $E_0$, and the field generated by all other charges in the system (the external field) be $E_0'$.
From the superposition principle, the total fields on either side of the boundary are:
$$E_2 = E_0' + E_0 = 2E$$
$$E_1 = E_0' - E_0 = E$$
By adding these two equations, we can find the external field $E_0'$:
$$2E_0' = 3E \implies E_0' = \frac{3}{2}E$$
The total force acting on this portion is:
$$\Delta F = \sigma \Delta S E_0'$$
Thus, the electrostatic pressure at the interface is:
$$P = \frac{\Delta F}{\Delta S} = \sigma E_0' = (\varepsilon_0 E)\left(\frac{3}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$

Similarly, for the second case ($E_1 = E$, $E_2 = -2E$), we have:
$$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (-2E) - \varepsilon_0 E = -3\varepsilon_0 E$$
The external field in this case is:
$$E_0' = \frac{E_1 + E_2}{2} = \frac{E - 2E}{2} = -\frac{1}{2}E$$
And the pressure is:
$$P = \sigma E_0' = (-3\varepsilon_0 E)\left(-\frac{1}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$

Why is the pressure the same?
We can obtain a general formula for the pressure $P$ using analogous reasoning. By substituting $E_0' = \frac{E_1 + E_2}{2}$ and $\sigma = \varepsilon_0(E_2 - E_1)$:
$$P = \sigma E_0' = \varepsilon_0(E_2 - E_1)\frac{E_1 + E_2}{2} = \frac{\varepsilon_0(E_2^2 - E_1^2)}{2}$$
From this final equation, we see that the electrostatic pressure depends on the squares of the electric fields. Therefore, changing the sign of $E_2$ leaves the pressure unchanged.