Let $O\in GL_n(\mathbb{C})$; if $A,B\in (\exp)^{-1}(O)$. Could we prove that there exists an unitary matrix, such that $$ A=UBU^\dagger + \Lambda, $$ in which, $\Lambda$ is a diagonal matrix whose entries are integer multiple of $2\pi$.
(thanks to @loup blanc, I have rephrased the question.)
A correctly written question is (for example):
Let $A\in GL_n(\mathbb{C})$; if $U,V\in (\exp)^{-1}(A)$, is it true that $U-V$ is a diagonal matrix ?
The answer is no, as the following example shows:
$A=I_2,U=0_2,V=\begin{pmatrix}0&-2\pi\\2\pi&0\end{pmatrix}$.
EDIT. I think you do not understand how the exponential works.
Proposition. Assume that $e^U=e^V$ and that $spectrum(U)$ is $2i\pi$ congruence-free (for every $s,t\in spectrum(U),s-t\notin 2i\pi \mathbb{Z}^*$). Then, there are $P\in GL_n$ and $\Lambda$ a diagonal matrix whose entries are integer multiple of $2iπ$ s.t. $U-V=P\Lambda P^{-1}$.
Proof. From a theorem by Hille, we deduce that $UV=VU$; then $e^{U-V}=I_n$; therefore $U-V$ is diagonalizable and its eigenvalues are in $2i\pi\mathbb{Z}$.
Remark. Proposition applies to my above example. Of course, if $U$ is not $2i\pi$ C.F., then the result is no more valid.