NOTE ON OBROUND VESSELS
While not a direct answer to your question, it should be quickly noted that ASME BPVC Section VIII, Mandatory Appendix 13, Section 10 has a section to design obround vessels (circular half sections with flat plate walls, see image below).
While not the same, these can typically optimize the available area compared to elliptical vessels (for a given dimension A and B). As such, these are almost always the preferred design choice compared to elliptical, and more commercially available. The equations, while handling a membrane stress discontinuity, are easier.
Thin wall assumption basis
We can easily derive a thin wall version of this. It can be noted that the ellipse shown above can be expressed as all points
$$(\frac{b}{2}\cos(\phi),\frac{a}{2}\sin(\phi))$$
with a parameter $\phi$ that varies from $0$ to $2\pi$. Note this isn't the same as the angle from the center, $\theta$, but is related to this angle by the relationship
$$\tan(\theta) = \frac{b}{a} \tan(\phi)$$
For a given $\phi$, the stress in a thin-wall vessel can be easily seen. Similar to a circular vessel, we take a cross section to a point at a specific $\phi$ through to the center. Using a similar thin wall approximation as cylindrical vessels, we presume the tangential (membrane) stress is uniform across this cut of a thickness $t$. We presume this is uniform regardless of $\phi$ as $t<<a ; t<<b$. The distance $D(\phi)$ is a function based upon $\phi$ from the edge, through the center to the other edge would be easily
$$D(\phi) = \sqrt{b^2\cos^2(\phi) + a^2\sin^2(\phi)}$$.
Since the wall is pushed uniformly across this cut with a pressure $P$, then it is clear that, similar to circles, for a load per unit length basis:
$$PD(\phi) = 2t\sigma(\phi)$$.
$$\sigma_t(\phi) = \frac{PD(\phi)}{2t}$$.
Presuming constant perimeter
Note that stress concentrations will definitely emerge, as the vessel will want to take on a cylindrical shape. Running a cut through both the x and y axis and running a statics analysis of this section, we note that a much smaller tangential stress comes from one axis than the other. The non-uniform strain will certainly cause bending moments in the thin wall material.
Without running a thick wall analysis, these stress concentrations cannot be determined exactly. For this thin wall presumption, we simply assume the perimeter will stay roughly the same before and after deflection. The perimeter of an ellipse is determined by using elliptical integrals. Defining:
$$k^2 = 1-\frac{a^2}{b^2}$$
Then the perimeter for the ellipse is $2bE(k)$, where $E$ is the elliptical integral of the second kind. We note that $k$ is a function of $b$. Taking the derivative of this with respect to $b$, and a lot of math, we find the beautiful result that:
$$\frac{E(k)}{F(k)} = \frac{a^2}{b^2}$$
We approximate $E(k)$ and $F(k)$, the elliptical integral of the first kind, as a constant for small changes in $k$ due to changes in $a$ and $b$. As such, for a small contraction of $b$, $\Delta b$, we find an expansion of $a$, $\Delta a$ as follows:
$$\Delta a = - \frac{b^3}{a^3} \Delta b$$
The cubic relation implies that highly eccentric ellipses will quickly snap from the elliptical shape into a cylindrical shape! As such, the above approximations will quickly become invalid. The negative implying the expansion is in converse to the contraction. With this information, we can attempt to resolve the stress concentrations to a reasonable approximation. The "radial" stress from a statics view indicates it should have a linear radial growth, while the perimeter analysis shows a cubic radial growth. This deflection would present itself as a force, adding a bending concentration at the tips of edge b (worst case) of approximately:
$$M = \frac{Pb^2(b^2-a^2)}{4a^2}$$
$$\sigma_b = \frac{3Pb^2(b^2-a^2)}{2t^2a^2}$$
This stress would have $b$ swapped with $a$ for the other axis, to obtain the two maximum stresses in the vessel. Note the original tangential stress will need to be added to the bending stress. As such, the total stress in the long-axis side would be:
$$\sigma_b = \frac{3Pb^2(b^2-a^2)}{2t^2a^2} + \frac{Pb}{2t}$$
Adding together with the axial stress of your case-by-case analysis (complicated, but could be assumed to be reasonable by dividing the area of an elliptical cap by the strange perimeter of an ellipse), and you have the stresses for a Von-Mises stress. This, as a function of the thickness, would be set to maximum value, and the minimum thickness would be the result that allows that Von-Mises stress to reach the maximum value. This portion is commonly utilized in pressure vessel design, as environmental loading adds more stress than your equation presumes.
I would run some tests to verify, but this is my conclusion as a PE for 8 years. Others can disagree as they see fit, and I'd be happy with a better analysis that shows better work.
Notes on thick wall vessel
I did attempt a thick wall version, but it quickly became apparent that I needed to invest more time than I was able to dedicate to it. Note you did not ask for this - and since your question and equation related to the thin-wall case, I saw no reason to push forward. Some notes, if you are familiar with elliptical integrals:
The practicalities are still an issue - obround works well even in thick-wall cases.
Sources are elusive for other works - Roark's was empty, as was ASME code. As such, I again defer to obround as the better method of construction for all practical applications.
