Suppose $x\geq 0$ is positive operator. Let $\sigma(x)$ denote the spectrum. If $0\in\sigma(x)$ denote $1_{\{0\}}$ to be the characteristic function. Is it true that $1_{\{0\}}(x)=0$? using the Borel functional calculus.
Denote $1_{\{0\}}(x)=a$. It is enough to show that $a=0$. I can see that for any polynomial $P$ holds $P(x)a=aP(x)=P(0)$. So $P(0)=0$ implies $P(x)a=aP(x)=0$. But I can't proceed further.
For any normal operator $x$, it is true that for any $z\in\sigma(x)$, $1 _{\{z\}}(x)=0$ if and only if $z$ is not an eigenvalue of $x$. The question therefore is, can we find a positive, non-invertible operator that is $1-1$? Yes, we can.
Consider for example $x:\ell^2\to\ell^2$ that maps $(a_n)\mapsto(a_n/n)$. It is obviously a bounded, linear and $1-1$ operator and it is also positive: we have $$\langle x(a_n)_n, (a_n)_n\rangle_{\ell^2}=\sum_{n}\frac{|a_n|^2}{n}\geq0$$ but this operator is not invertible, since it is not bounded below: the canonical orthonormal basis $(e_n)$ of $\ell^2$ is a sequence of unit vectors such that $x(e_n)\to0$.
Since we can find such an operator, that means that what you are asking is not necessarily true.