I found the following assertion and I would like to know why does it hold:
Let $\pi$ be a representation of a von-Neumann algebra such that $\pi=\pi_1 \oplus \pi_2$ where $\pi_1,\pi_2$ are irreducible representation. and let {$x_1,x_2$} linearly independent. Now by theorem 1 from here I know there is a self adjoint operator $B$ such that $Bx_i=(-1)^ix_i \ \ (i=1,2)$ and by taking exponent of fitting operator we can find unitary operator $A$ such that $Ax_i=(-1)^ix_i \ \ (i=1,2)$.
My Question is if I can find now a unitary element $a$ of the algebra such that $\pi(a)x_i=(-1)^ix_i \ \ (i=1,2)$. In other words any such unitary operator can be written as $\pi(a)$ where $a$ is unitary element of the algebra.
Note that $A$ is also unitary operator in $\pi(M)$ (where $M$ is the algebra).
If I understood your question correctly then the answer is yes only in case $\sigma(A)\neq S_1$ (that is the unit sphere in $\mathbb{C}$).
Indeed in that case we can find a continuous function $f$ on $\sigma(A)\subsetneq S_1$ such that $f(A)$ is self-adjoint and $e^{if(A)}=A$ (try to figure out what is $f$ by yourself). Notice that $f(A)\in \pi(M)$ so if $f(A)=\pi(X)$ for $X\in M$ we have $f(A)=\pi(Y)$ where $Y=\frac{1}{2}(X+X^*)$ a self-adjoint element. So by taking the unitary element $a=e^{iY}$ we end up with $\pi(a) = \pi(e^{iY}) = e^{i\pi(Y)} = e^{if(A)} = A$ as wanted.//