Let $A$ be a unital $C^*$-algebra and $a ∈ A$ arbitrary. I have already shown that for $a ∈ \text{Inv}(A)$, we have $|a| := \sqrt{a^*a} ∈ \text{Inv}(A)$, so we can indeed, for $a ∈ \text{Inv}(A)$, define $u := a|a|^{-1}$ so that $a = u|a|$. (I see that we are essentially 'normalising' the element $a$, insofar as that is meaningful here.)
Now we can just compute:
$$u^{-1} = |a|a^{-1} = \sqrt{a^*a}a^{-1} = (a^*)^{\frac{1}{2}}a^{\frac{1}{2}}a^{-1} = (a^*)^{\frac{1}{2}}a^{-\frac{1}{2}} = \sqrt{a^*a^{-1}} $$ and $$u^* = (a|a|^{-1})^* = (|a|^{-1})^*a^* = \left(\left(\sqrt{a^*a}\right)^{-1}\right)^*a^* = \left((a^*a)^{-\frac{1}{2}}\right)^*a^* = (a^*a)^{-\frac{1}{2}}a^* = a^{-\frac{1}{2}}(a^*)^{\frac{1}{2}} = (a^{-1})^{\frac{1}{2}}(a^*)^{\frac{1}{2}} = \sqrt{a^{-1}a^*} $$
So apparently, if $u$ is to be unitary (i.e., $u^* = u^{-1}$), we must have that $a^{-1}$ commutes with $a^*$, but I see no reason to assume that.
Yes, $u:=|a|a^{-1}$ is unitary. To prove it one computes $$ u^*u = (a^*)^{-1}|a|^2a^{-1}= $$$$=(a^*)^{-1}a^*aa^{-1}= 1, $$ while $$ uu^* = |a|a^{-1}(a^*)^{-1}|a|= $$$$ =|a|(a^*a)^{-1}|a| =|a||a|^{-2}|a|=1.$$