This can be just a typo, but you have a wrong parametrization of the sphere, it should be
$$
\begin{align}
x & = a r \sin \theta \cos \varphi, \\
y & = b r \sin \theta \sin \varphi, \\
z & = c r \cos \theta.
\end{align}
$$
Your limits for each variable are correct though. Your Jacobian is incorrect because you forgot to take in account the factors $a,b,c$. It should be
$$\frac{\partial (x,y,z)}{\partial (r, \theta, \varphi)} = - abc r^2 \sin \theta.$$
The $-$ sign is because this parametrization of the sphere reverses orientation.
When I edited your post I made sure to clarify some things but I didn't edit a couple of mistakes, which I intend to explain now:
1) I turned $d$'s into $\partial$'s for the Jacobian to correct your notation.
2) The notations $d(x,y,z)$ and $d(r, \theta, \varphi)$ don't make sense, it is best to stick to $dx \, dy \, dz$ and $dr \, d \theta \, d \varphi$.
Your set up for the mass is correct if you fix the Jacobian and add $abc$. The calculation seems to be too (I haven't checked that thoroughly).
I don't understand what you mean by $x_s$. If you want to compute the $x$ coordinate of the center of mass, I assume you are using
$$x_s = \frac{1}{M} \int\limits_{E} x \mu \, dV, \text{ or } x_s M = \int\limits_{E} x \mu \, dV.$$
As you have seen, this is zero, just like the others will be. This has to do with mjqxxxx's comment that the ellipsoid has symmetry about all axis, therefore its center of mass has to be at the origin.
Hint:
It seems that all what you need is the principle of inertia wrt to a translation of the axes:
The moment of inertia of a body wrt to a given line is equal to the moment wrt that line of the mass of the whole body concentrated on the barycenter, plus the moment of inertia of the body around the line when parallel translated into the barycenter.
I found the reference (official naming in english) you requested, that's called Parallel axis theorem.
-- addendum --
If $J$ indicates the moments in the base reference system
(origin in the geometric center) and $I$ those with the reference
translated at the barycebter $(x_b, y_b, z_b)$ then for instance
$$
\eqalign{
& J_{\,x\,y} = \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt V}
{xy\rho (x,y,z)dV} = \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt V}
{\left( {x_b + \Delta x} \right)\left( {y_b + \Delta y} \right)\rho (x,y,z)dV} = \cr
& = \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt V}
{\left( {x_b y_b + x_b \Delta y + y_b \Delta x + \Delta x\Delta y} \right)\rho (x,y,z)dV} = \cr
& = x_b y_b \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt V}
{\rho (x,y,z)dV} + 0 + 0 + \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt V}
{\Delta x\Delta y\rho (x,y,z)dV} = \cr
& = x_b y_b M + I_{\,x\,y} \cr}
$$
So, once you have the matrix, i.e. the integrals, in the base reference, and the position of the barycenter in that reference,
it's quite easy to get the matrix in the reference having the barycenter as origin.
Best Answer
Your conjecture is correct:
The full ellipsoid has moment of inertia around the $x$-axis $$ I_x=\iiint\limits_{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le 1}\rho\,(z^2+y^2)\,dx\,dy\,dz $$ which is known to be $I_x=\frac{M}{5}(b^2+c^2)$ where $M$ is the mass and $\rho$ the uniform density $$\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}abc}\,.$$
If we denote by $I^+_x,I^-_x$ the moments of the two half ellipsoids with $z\ge 0\,,$ resp. $z\le 0$ we know from symmetry that they must be equal and, by the integral formla, add up to $I_x\,.$
The fact that the off diagonal elements $I^\pm_{xy}\,,I^\pm_{xz}\,,I^\pm_{yz}$ are zero is easy to see. For example, $$ I^+_{xy}=\iiint\limits_{\scriptstyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le 1\atop\scriptstyle z\ge 0}-\rho\,xy\,dx\,dy\,dz $$ is zero because the integral $\int_{-d}^dxy\,dx$ is zero and $$ d=a\sqrt{1-\frac{y^2}{b^2}-\frac{z^2}{c^2}}\,. $$