Proposition 1.4 in Takesaki's book "Theory of operator algebra": about the spectral integral

Consider the following fragment from Takesaki's book "Theory of operator algebra I":

Can someone explain why the restriction $x\vert_{e_n(\mathscr{H})}$ is invertible? Intuitively, I would suspect that it has to do something with the fact that we cut out $0$ from the spectrum somehow, but I can't make this precise.

I also get a feeling that making use of the fact that we have a unital isometric $*$-isomorphism $B^\infty(\sigma(x)) \to B(H): f \mapsto f(x) = \int f(\lambda)de(\lambda)$ seems relevant here.

Any help will be highly appreciated!

Since $$xe_n=\int_{|\lambda|\geq1/n}\lambda\,de(\lambda)\geq \tfrac1n\,\int_{|\lambda|\geq1/n}\,de(\lambda)=\tfrac1n\,e_n,$$ when considered as an operator on $e_n \mathfrak H$, it satisfies $$ x\geq \tfrac1n\,I. $$ This means that $x-\tfrac1n I$ is positive, which implies that $\sigma(x)\subset\big[\tfrac1n,\infty\big]$.