Actually, the factor of $2$ is unimportant. The center of mass in the $y$ direction is given by
$$\bar{y} = \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)(r \sin{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)}$$
Note that the weight of the $y$ coordinate is expressed in the $r \sin{\theta}$ term in the numerator. Note also the boundary of the lamina is expressed in its polar form, $r=\sin{\theta}$. Evaluating the radial integrals, we get
$$\bar{y} = \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \sin^5{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}$$
or
$$\bar{y} = \frac{6}{5}$$
For $\bar{x}$, we do a similar calculation:
$$\begin{align}\bar{x} &= \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)(r \cos{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)}\\ &= \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \cos{\theta} \sin^4{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}\\ &= \frac{9}{20}\end{align}$$
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.
Best Answer
See the picture below.
As you can see, $\cos x$ is above $\sin x$ or $\cos x>\sin x$ for $\left[0,\dfrac\pi4\right]$. Therefore $$ \begin{align} M&=\int_0^\frac\pi4(\cos x-\sin x)\ dx=\sqrt{2}-1,\\\\ M_y&=\int_0^\frac\pi4x(\cos x-\sin x)\ dx=\frac14(\sqrt{2}\pi-4),\quad\Rightarrow\quad\text{use IBP} \end{align} $$ and $$ \begin{align} M_x&=\frac12\int_0^\frac\pi4(\cos x+\sin x)(\cos x-\sin x)\ dx\\ &=\frac12\int_0^\frac\pi4(\cos^2 x-\sin^2 x)\ dx\\ &=\frac12\int_0^\frac\pi4\cos2x\ dx\\ &=\frac14. \end{align} $$ I think you can handle it from here. I hope this helps.