The following Euler Equation
$$ L[y] = x^2y''+\alpha xy'+\beta y = 0$$
has the solutions of the form
$$ y= c_1 x^{\frac{1}{2}\left( \sqrt{(\alpha-1)^2-4\beta}~-~(\alpha-1)\right)}+c_2x^{\frac{1}{2}\left(-\sqrt{(\alpha-1)^2-4\beta}~-~(\alpha-1)\right)}$$
The solution is only defined for $x>0$ which allows the solution to be defined at fractional exponents.
However when the equation is
$$ L[y] = (x-x_0)^2y''+\alpha (x-x_0)y'+\beta y = 0$$ the solution becomes
$$ y= c_1 |x-x_0|^{\frac{1}{2}\left( \sqrt{(\alpha-1)^2-4\beta}~-~(\alpha-1)\right)}+c_2|x-x_0|^{\frac{1}{2}\left(- \sqrt{(\alpha-1)^2-4\beta}~-~(\alpha-1)\right)}$$
making the solution defined for all $x$.
Why can't the solution for the first equation have $|x|$ instead of $x$ making it defined for all values of $x$?
It can. There is no difference between the two cases.
The equation itself is singular at $x=x_0$, so there is no real connection between a solution for $x > x_0$ and for $x < x_0$. It is just for convenience that we might write the equation in terms of $|x-x_0|$ rather than using separate formulas with powers of $x - x_0$ for $x > x_0$ and of $-x + x_0$ for $x < x_0$.
On the other hand, if you consider the equation for complex $x$, the formulas with absolute values would be wrong, because they would not be differentiable.