Let $A$ be a unital algebra and $B$ a subalgebra such that $B+\mathbb C 1=A$. Suppose $1$ is the multiplicative identity of $A$ while $e$ is is the multiplicative identity of $B$. I want to prove the following:
When $\lambda \in \mathbb C-\{0\}$, then the invertibility of $b+\lambda1$ in $A$ is equivalent to the invertibility of $b+\lambda e$ in $B$.
In other words,
$(b+\lambda e)b_B=e$ for some $b_B\in B$ if and only if $(b+\lambda 1)b_A=1$ for some $b_A\in A$ .
I find either direction is hard for me. I don't know if $A$ and $B$ must be assumed to be $C^*-$algebra, this is from Murphy's book, page 45 on the top.
There is no need to assume that $A$ nor $B$ are C$^*$-algebras. Just algebras.
If $b+\lambda e$ is invertible in $B$, then there exists $c\in B$ with $$c(b+\lambda e)=(b+\lambda e)c=e.$$ Let $d=c+\lambda^{-1}(1-e)\in A$. Note that $x(1-e)=(1-e)x=0$ for all $x\in B$. Then \begin{align} d(b+\lambda 1)&=c(b+\lambda 1)+\lambda^{-1}(1-e)(b+\lambda 1) =cb+\lambda c+(1-e)\\ \ \\ &=cb+\lambda ce+1-e =c(b+\lambda e)+1-e\\ \ \\ &=e+1-e=1 \end{align} (and a similar computation for $(b+\lambda 1)d$). So $b+\lambda e$ is invertible in $A$.
Conversely, if $b+\lambda 1$ is invertible in $A$, there exists $d\in A$ with $d(b+\lambda 1)=1$ Then $$ (ede)(b+\lambda e)=ede(be+\lambda e)=ed(b+\lambda 1)e=e $$ (and similar computation for $(b+\lambda e)(ede)$). So $b+\lambda e$ is invertible in $B$. Note that $B+\lambda\mathbb C 1=A$ implies that $B$ is an ideal, so $ede\in B$.