My first question is, given two different elements $L,R$ in the Lie group $SU(N)$ can one always make write $$L=V A, \quad \quad R=VA^\dagger,$$ for some $V, A \in SU(N)$ ?
Secondly, how does this generalise to general Lie (or even non-Lie) groups. Is it always possible to decompose in the above fashion (with $A^\dagger$ generalised to $A^{-1}$)?
I first claim that $A$ and $V$ exist iff $R^{-1}L$ is the square of some element in the group.
First, if $A$ and $V$ exist, then $R^{-1}L = AV^{-1} VA = A^2$ is a square. Conversely, if $R^{-1}L$ is a square, then there is a solution to $A$ to the equation $A^2 = R^{-1} L$. Setting $V = LA^{-1}$ works. That is, $VA = LAA^{-1} = L$ while $VA^{-1} = L(A^2)^{-1} = L(R^{-1}L)^{-1} = R$.
Having established the claim, note that in any connected compact Lie group $G$, the exponential map is surjective. Given $g\in G$, we may write $g = \exp(X)$ for some $X$ in the Lie algebra of $G$. Then $\exp(1/2 X)^2 = \exp(X) = g$, so $g$ is a square. Hence, in a connected compact Lie group, you can always find $V$ and $A$.
On the other hand, in some groups, not every element is a square. The smallest is $\mathbb{Z}/2\mathbb{Z}$, where the non-trivial element is not a square.