Is the statement "Projection on kernel of $D\in\tilde A$ is element of $A$" correct or an error in the paper? I am probably missing something, but there seems to be a (relevant) error in this paper page 351, the statement in question being:

Then $D$ belongs to the algebra $\widetilde{A\rtimes_r\Gamma}$($\subset B(\ell^2(\Gamma\times\mathbb N))$) obtained by adjoining a unit to $A\rtimes_r\Gamma$, the point $0$ is isolated in its spectrum, and the orthogonal projection onto the kernel of $D$, which is by spectral theory ($D$ is positive) an element $p$ of $A\rtimes_r\Gamma\subset\ell^\infty(\mathbb N;\mathcal K(\ell^2(\Gamma)))$.

I think we can not have $p\in A\rtimes_r\Gamma$. My understanding is that the situation above is this (replacing $A\rtimes_r\Gamma$ with just $A$):

  
*
  
*$A\subset B(H)$ a nonunital $C^*$-subalgebra for a Hilbert space $H$,
  
*$\tilde A=C^*(A,1)\subset B(H)$,
  
*$0\le D\in\tilde A$ with $0$ isolated in the spectrum of $D$,
  
*$p\in B(H)$ the projection on $\ker(D)$,
  
*and (crucially) $p\in A$.
  

But by continuous functional calculus we get $p=1_{\{0\}}(D)\in\tilde A\setminus A$, where $1_{\{0\}}$ is the characteristic function of $0\in\sigma(D)$, which is continuous. This clearly contradicts $p\in A$, where is the mistake?

I can't verify the results in the paper, but I think I can show why your objection doesn't work.
Let $X = \{0\} \cup [1, 2]$.  To match your notation, set $A = \{f \in C(X) : f(1) = 0\}$, so that $C^*(A, 1) = \tilde A = C(X)$.  Finally, set $D = x$, the identity function.  Note that $0 \le D \in \tilde A\setminus A$.  Then $p = 1_{\{0\}}(D) = 1_{\{0\}} \in A$.
There's no contradiction here.  It's possible for $1_{\{0\}}(D)$ to belong to $A$.  It just can't belong to $C^*(D)$.
Of course, the paper would still have a problem if they said that $D \in A$, since then $1_{\tilde A} = 1_{\{0\}}(D) + 1_{(0, \infty)}(D) \in A + C^*(D) \subset A$, and $A$ is non-unital.  But I think everything works out as long as $D$ (and in particular its complementary spectral projection $1_{(0, \infty)}(D)$) is outside of $A$.
