Let the ellipse
$$\frac{x^2}{a^2}+\frac{(y-b)^2}{b^2}=1\quad(1)$$
and the semicircle
$$y=\sqrt{1-x^2}\quad(2)$$
The ellipse is clearly tangent to $x$-axis.
If we substitute the semicircle equation in the ellipse equation, we get:
$$(a^2-b^2)y^2-2a^2by+b^2=0 \quad(3)$$
As the semicircle and the ellipse are tangent the discriminant of equation $(3)$ must be zero.
Therefore
$$b=\sqrt{a^2(1-a^2)} \quad(4)$$
But we know that the ellipse area can be calculated by:
$$A=\pi a b \quad(5)$$
From $(4)$ and $(5)$, we get:
$$A= \pi \sqrt{a^4(1-a^2)} \quad (6)$$
To calculate the maximum of $A$ we can use the AM-GM inequality:
$$(\frac{a^2}{2}\frac{a^2}{2}(1-a^2))^{\frac{1}{3}} \leq \frac{\frac{a^2}{2}+\frac{a^2}{2}+(1-a^2)}{3}=\frac{1}{3} \Rightarrow$$
$$a^2a^2(1-a^2)\leq \frac{4}{27}. \quad(7)$$
We know that the equality holds for
$$\frac{a^2}{2}=(1-a^2).\quad(8) $$
Therefore
$$a=\sqrt{{\frac{2}{3}}}\quad(9)$$
From $(4)$ and $(9)$, we get:
$$b=\frac{\sqrt{2}}{3}$$
Finally we get
$$e=\sqrt{\frac{2}{3}}.$$
This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.
First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$\phi:P\subset\mathbb{R}^2\to\mathbb{R}^3,\qquad \|d\phi^Td\phi\|_2 \leq 1$$
of the piece of paper $P$ into $\mathbb{R}^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $\phi$ sends the paper boundary $\partial P$ to a curve $\gamma$ in the plane.
We can thus partition your problem into two pieces: (I) given a fixed curve $\gamma$, what is the volume of the volume-maximizing surface $M_{\gamma}$ with $\phi(\partial P) = \gamma$? (II) Can we characterize $\gamma$ for which $M_{\gamma}$ has maximum volume?
Let's consider the case where $\gamma$ is given. We can partition $M_{\gamma}$ into
1) regions of pure tension, where $d\phi^Td\phi = I$; in these regions $M_{\gamma}$ is, by definition, developable;
2) regions where one direction is in tension and one in compression, $\|d\phi^Td\phi\|_2 = 1$ but $\det d\phi^Td\phi < 1$.
We need not consider $\|d\phi^Td\phi\|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $\phi$ a short map.
Let us look at the regions of type (2). We can trace on these regions a family of curves $\tau$ along which $\phi$ is an isometry. Since $M_{\gamma}$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_{\gamma}$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $\tau$. In other words, for some stress $\sigma$ constant along each $\tau$, at all points $\tau(s)$ along $\tau$ we have
$$\hat{n} = \sigma \tau''(s)$$
where $\hat{n}$ the surface normal; it follows that (1) the $\tau$ follow geodesics on $M_{\gamma}$, (2) each $\tau$ has constant curvature.
The only thing I can say about problem (II) is that for the optimal $\gamma$, the surface $M_\gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $\approx 1.236$ liters, less than Joriki's solution).
I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $\gamma$ and $M_{\gamma}$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper
Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).
EDIT 2: A sketch of the orientation of $\tau$ on the surface:
Best Answer
The isoperimetric inequality says that the largest volume enclosed by a surface of given area is the volume of the sphere of this area. In the linked problem we do not want to "enclose" fluid, but just "hold" it. One then has to prove that the largest volume that can be "hold" in a surface of given area is the volume of a half ball whose spherical boundary part has the given area. Given that we can argue as follows:
The A4 sheet has area $A=210\cdot297$ mm$^2$ If we cut up this sheet into tiny strips and glue the strips together to a half sphere of radius $r$ we obtain $2\pi r^2<A$, or $r<99.63$ mm. The volume of the resulting half ball then is ${2\pi\over3}r^3<2.071$ liter.