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.
Here's your answer! Hope it helps.
Best Answer
Hint.
The answer of Matthew Leingang is general. In your case the region is a triangle so you can find the centroid (that coincides with the center of mass since the density is constant) using the fact that the coordinates of the centroid are the means of the coordinates of the vertices.