Does regularity of distributions have anything to do with definiteness of their product?

What is going on here is that the example you give of $\frac{1}{\sqrt{x}}$ is not regular. The singular support is not empty, it is equal to $\{0\}$. So the theorem you mentioned does not apply. You trivially get an element of $\mathcal{D}'(\mathbb{R}\backslash\{0\})$ but you still have to work harder in order to get a distribution on the whole real line.

I think you are referring to the definition of product of distributions due to Hoermander based on the notion of wavefront set. The corollary you mention has this precise form: if the sigular supports of a pair of distributions have empty intersection, then their (Hoermander) product is well defined.

The singular support of the distribution $u$ is the complement of the union of the open sets $U$ such that $u(f) = \int g_U f dx$ for some $C^\infty$ function $g_U$ and every test function supported in $U$.

With the said definition, the singular support of $u= 1/\sqrt{x}$ is $0$ (though I do not understand well how you define $u$ for $x<0$). So the corollary does not apply.