Let $H$ be a infinite dimensional Hilbert Space. $A \in B(H)$ and $A$ self-adjoint. Let $\sigma_e(A)$ be the essential spectrum of $A$. Since $A$ is self-adjoint, we can assume the convex hull of $\sigma_e(A)$ is $[a, b]\,(a, b \in \mathbb{R})$. Define $f$ to be $$f(x) = \begin{cases} a, & x \leq a\\x, & x \in [a, b]\\b, &x \geq b \end{cases}$$
Let $\mathcal{K}$ be the ideal of compact operators and let $\pi$ be the canonical mapping from $B(H)$ to $B(H)\,/\,\mathcal{K}$. Since $\sigma_e(A) = \sigma(\pi(A))$ we have $\pi(f(A)) = f(\pi(A)) = \pi(A)$.
Let $E_A(\Delta) = \chi_{\Delta}(A) \in W^*(A)$. Then how to show that $E_A(b - \frac{1}{n}, b)$ has infinite rank for all $n \in \mathbb{N}$. If $v$ is a unit vector in the range of $E_A(b - \frac{1}{n}, b]$, how can I show $\langle Av, v\rangle > b - \frac{1}{n}$? I can only tell $b \in \sigma_e(A)$ and hence $Ker[A - b]$ has infinite rank but $b \in (b - \frac{1}{n}, b]$ might be one of the reasons. In general for $f \in L^{\infty}[\sigma(A)]$ spectral mapping property might not hold and for the same $f$, I am not sure if $f[\pi(A)] = \pi[f(A)]$.
This question is inspired by Lemma II.5.1 in $C^*$-Algebras by Example. Below is the lemma
Lemma II.5.1 Let $\phi$ be a state on a separable $C^*$-subalgebra $\mathfrak{A}$ os $B(H)$ with the property that $\phi(\mathfrak{A}\,\bigcap\,\mathcal{K}) = 0$. Then show that there is a sequence of unit vectors $x_n$ converging weakly to $0$ such that the corresponding vector states $\psi(A) = \langle A\,x_n, x_n\,\rangle$ on $\mathfrak{A}$ converge weak-$\ast$ (i.e. pointwise) to $\phi$
Inside the proof, the author defines $\mathcal{S}_e$ to be the set $\{\,\psi\,\text{states on } \mathfrak{A}\,\vert\,\exists\,\{x_n\}_{n \geq 1} \subseteq H_{\leq 1}\,\text{such that}\,\langle A\,x_n, x_n \rangle \rightarrow \psi(A)\}$. After fixing a self-adjoint element $A \in \mathfrak{A}$, the author then tries to show $W_e(A) = \{\psi(A)\,\vert\,\psi \in \mathcal{S}_e\} = [a, b]$ and directly claims $E_A(b - \frac{1}{n}, b]$ has infinite rank.
If in this case if $\sigma_e[f(A)] = f[\sigma_e(A)]$ for some $f \in L^{\infty}[\sigma(A)]$ then the title can be answered. I wonder if this is true in general. For example, given $A$ normal, what other properties (hopefully not trivial) does $A$ need to have for $\sigma_e[f(A)] = f[\sigma_e(A)]\,\forall\,f \in L^{\infty}[\sigma(A)]$? If there is some general conditions that makes $\sigma_e[f(A)] = f[\sigma_e(A)]$ true for a self-adjoint (normal resp.) $A$ and for some $f \in L^{\infty}[\sigma(A)]$, what properties does the set of these functions have?
If $E(b-\tfrac1n,b]$ has finite-rank, then you can write $A=AE(-\infty,b-\tfrac1n]+AE(b-\tfrac1n,b]$. Now apply $\pi$, the quotient map onto the Calkin algebra, and you get $\pi(A)=\pi(AE(-\infty,b-\tfrac1n])$, implying that $\sigma_e(A)\subset[a,b-\tfrac1n]$, contradicting that the convex hull of $\sigma_e(A)$ is $[a,b]$.
As for your general question, the Spectral Mapping Theorem holds for the Borel functional calculus (and, in particular, in the Calkin algebra and its double commutant; recall that the spectrum does not depend on the algebra). See for instance Theorem IX.8.11 in Conway's A Course in Functional Analysis.