Let $A$ be a unital $C^\ast$ algebra. It is stated in this book that for any $C^\ast$ subalgebra we have $\sigma_B(b)\cup\{0\} = \sigma_A(b)\cup\{0\}$.
The reasoning why this should be true is this: First, if $B$ contains $1_A$ then it is a theorem that $\sigma_A(b) = \sigma_B(b)$. Next (p 44), it is noted that if $B$ does not contain a unit and $B + \mathbb C= A$ then $\sigma_A(b) = \sigma_B(b)$. Finally (p 45) it is noted that if $1_B \neq 1_A$ then $\sigma_A(b) = \sigma_B(b)\cup \{0\}$.
How does it follow that if $B$ is a very special subalgebra with the property that $B + \mathbb C=A$ that then also for any subalgebra we have equality of spectra? I believe this does not follow.
Take any $B$. Then $B+\mathbb C\,1_A$ is a C$^*$-subalgebra of $A$ that contains $1_A$, so $\sigma_{B+\mathbb C\,1_A}(b)=\sigma_A(b)$.
Now by page 44, applied to the inclusion $B\subset B+\mathbb C\,1_A$, you have $\sigma_B(b)=\sigma_{B+\mathbb C\,1_A}(b)$, and you are done.