(I tried to post this as a comment but it mangled the equation.)
Even for a degenerate $E_1 = \{(x,0): -1\le x\le1\}$, the implicit form of $E_2$ is already pretty complicated. Mathematica gives me, after some massaging,
$$\begin{align}&\int_{-1}^1 \sqrt{(x-t)^2 + y^2}\ \mathrm dt \\
&\qquad = (1-x)\sqrt{(1-x)^2+y^2}+(1+x)\sqrt{(1+x)^2+y^2} \\
&\qquad +\ y^2\log\left(\frac{\left(1-x+\sqrt{(1-x)^2+y^2}\right)\left(1+x+\sqrt{(1+x)^2+y^2}\right)}{y^2}\right)\end{align}$$
which you want the level sets of. This is for the restriction to the $xy$-plane, but since there is rotational symmetry in this case, one can just replace $y^2$ with $y^2+z^2$ for the full 3D surface.
When $E_1$ is nondegenerate and the arc length of the ellipse comes into play, the equation will probably get much worse.
Given five points of the conic, and using Pascal's theorem you are able to find the intersection of the conic with an arbitrary line that contains one of these. Moreover, you are able to find the tangent to the conic at any of these points.
Now, you are able to produce pairs of conjugate diameters that intersect at the center of the conic (ellipse). To do that, notice that the middle of the chords parallel to a given direction are all on the line passing through the center of the ellipse (the conjugate diameter). This produces conjugate directions, to get the sizes of the diameters one uses Prop XXII from the source below. Up to now, we haven't mentioned angles.
The rest, and most interesting part is explained in an old book that you can find here: An elementary treatise on the Geometry of Conics, page 96, ex 8. There îs one beautiful basic fact being used , that the product of the intercepts on the tangent to an ellipse by a pair of conjugate directions is constant (equals the square of the half the diameter parallel to the tangent). With this and some classical construction one is able (ex 8 mentioned above) to construct a pair of conjugate directions that are perpendicular, so the axes. One now uses a consequence of prop XXII in the same book to get the sizes of the diameters.
I noticed now that one can also use results from page 106 of the same book (ex 14 or 15).
Also see Rytz's construction.
All these results seem classical, to be known by Newton and maybe even before him (Apollonius?), but forgotten these days, since the geometry of conics is nowadays seen as a particular case of "analytic geometry", excepts perhaps the "fun things" like Pascal, Brianchon, Poncelet and very basic things.
Best Answer
I am interpreting the question as asking for a minimum number of points that determine a unique ellipse passing through them.
Four points is not enough. The two ellipses $$ x^2+2y^2=3\qquad\text{and}\qquad 2x^2+y^2=3 $$ both pass through the points $(x,y)=(\pm1,\pm1)$ (all four sign combinations).
Five points $P_j=(x_j,y_j), j=1,2,3,4,5,$ (in general) does suffice. But we get more than we bargained for. Namely five points determine a conic. The general quadratic equation $$ a_1x^2+a_2xy+a_3y^2+a_4x+a_5y+a_6=0\qquad(*) $$ has six unknown coefficients. By plugging in the coordinates of the points $P_j$ we get a homogeneous system of five equations in the six unknowns. Linear algebra then tells us that the system has a non-trivial solution. Usually the solution is determined up to a scalar multiple, but scaling all the coefficients in $(*)$ won't change the curve, so we can ignore that.
Two caveats: