Consider the Hilbert space $l^2(\mathbb{Z})\otimes l^2(\mathbb{N})$. Let $p$ be a rank-$1$ projection on $l^2(\mathbb{N})$, and let $$P=1\otimes p.$$ Let $S$ be the right-shift operator on $l^2(\mathbb{Z})$.
Question: Is the operator $1+P+PS$ unitary (or even invertible)?
Comment: So far I've calculated that $$(1-P+PS)(1-P+PS)^*=(1-P+PS)(1-P+S^*P)=1+(PS(1-P))^*+PS(1-P),$$ but this doesn't seem to reduce to $1$.
Assuming you mean $S\otimes 1$ when you write $S$ in your compuations then yes, your operator is a unitary. This follows from the fact that $S$ is a unitary and that the factors commute. In your computation, the second and third terms are $$ (1\otimes p)(S\otimes 1)(1\otimes (1-p))=S\otimes p(1-p)=0 $$ and its adjoint. So $$ (1-P+PS)(1-P+PS)^*=1. $$ On the other side, $$ (1-P+PS)^*(1-P+PS)=1-P+S^*PS=1-P+PS^*S=1. $$