I don't know where the Wikipedia article is coming from, but I don't think what you want to do can be done.
First, I'm not aware of a context where a differential operator is bounded as an operator on a Hilbert space. So I don't see how you would put a differential operator inside a C$^*$-algebra.
Second, differential operators are not positive, so doing functional calculus with a function that is only defined for positive values (as are fractional powers) is no-go.
Before tackling the question itself, it is perhaps useful to discuss a minor point regarding the fact that the unit of
$p A p$ is $p$, rather than $I$. To highlight this difference,
whenever we are given an element
$b\in p A p$,
we will write $\sigma _p(b)$ for the spectrum of $b$ relative to
$p A p$,
reserving the notation
$\sigma (a)$
for the spectrum of any element $a\in A$ relative to $A$ (or, equivalently, to $B(H)$).
Leaving aside the trivial case in which $p=1$, observe that no element $b\in pAp$ is invertible relative to $A$, so 0 is always in $\sigma (b)$. In
fact it is easy to show that, for every such $b$, one has
$$
\sigma (b) = \sigma _p(b)\cup \{0\}.
\tag{*}
$$
Likewise, if $b\in p A p$, and $f$ is a holomorphic function on a neighborhood of $\sigma _p(b)$, we will denote by
$f_p(b)$
the outcome of the holomorphic functional calculus computed relative to $pAp$. As before, we will reserve the
undecorated expression $f(a)$ for the holomorphic functional calculus relative to $A$.
In the event that $f$ is holomorphic on the larger set $\sigma _p(b)\cup \{0\}$, one may easily prove that
$$
f(b)=f_p(b)+f(0)(1-p),
\tag{**}
$$
for every $b\in p A p$.
This said, let $b\in \mathscr A$ and let $f$ be a holomorphic function on an open set $U$ such that $\sigma _p(b)\subseteq U$.
CASE 1: Assuming first that $0\in U$, we have by ($*$)
that $f$ is
also holomorphic on a neighborhood of $\sigma (b)$ and we have by
hypothesis that $f(b)\in \mathscr A$. Applying ($**$) it then follows that
$$
f_p(b)=pf(b)p \in p\mathscr A p,
$$
as desired.
CASE 2: $0\notin U$.
In this case it is clear that $0\notin \sigma _p(b)$, and
since $\sigma _p(b)$ is compact, one has that $r:=\text{dist}(0,\sigma _p(b))>0$. So
$$
\sigma _p(b)\subseteq V:= \mathbb C\setminus \overline {B_{r/2}(0)}.
$$
By restricting $f$ to $U\cap V$, we may assume that $U\subseteq V$. The open ball $B_{r/3}(0)$ is therefore disjoint from
$U$, so we may extend $f$ to $U\cup B_{r/3}(0)$ by declaring it to be identically zero on $B_{r/3}(0)$. Consequently the extended $f$
is now defined on a neighborhood of $\sigma _p(b)\cup \{0\}$.
Since the extension process did not change the values of $f$ on points of $\sigma _p(b)$, the outcome of $f_p(b)$ remains
unchanged and hence the conclusion follows as in case 1.
Best Answer
If you have a bounded operator $A$, then the holomorphic functional calculus is always an option, and it is based on Cauchy's integral representation: $$ f(A) = \frac{1}{2\pi i} \oint_{C} f(\lambda)``\frac{1}{\lambda I-A}"\,d\lambda = \frac{1}{2\pi i} \oint_{C} f(\lambda)(\lambda I-A)^{-1}\,d\lambda. $$ The contour $C$ is any simple closed rectifiable curve enclosing the spectrum $\sigma(A)$ in its interior, and $f$ is a holomorphic function on a neighborhood of $C$ and its interior. For all such functions $f$, $g$, $f(A)g(A)=(fg)(A)$. That is $f\mapsto f(A)$ is multiplicative and linear. You cannot extend this calculus to functions $f$ which are continuous, at least not in general. However, if $A$ is normal or selfadjoint, then the calculus does extend to continuous functions.
You can see why the holomorphic functional calculus fails to extend to general continuous functions: the resolvent $(\lambda I-A)^{-1}$ may have a pole of order $n > 1$. For example, if $N$ is a nilpotent matrix of order $n \ge 2$, then the resolvent is $$ \frac{1}{\lambda I - N} = \frac{1}{\lambda}\frac{1}{I-\frac{1}{\lambda}A}=\frac{1}{\lambda}\left[I+\frac{1}{\lambda}N+\frac{1}{\lambda^{2}}N^{2}+\cdots+\frac{1}{\lambda^{n-1}}N^{n-1}\right] $$ Then the holomorphic functional calculus gives $$ f(N) = \frac{f(0)}{0!}I+\frac{f'(0)}{1!}N+\cdots+\frac{f^{(n-1)}(0)}{(n-1)!}N^{n-1}. $$ So it is impossible for the functional calculus to extend to general continuous functions for such $N$. Nilpotent matrices arise naturally when studying even a general matrix, as can be seen from the Jordan Canonical form. In fact, you can derive the Jordan form from the holomorphic functional calculus. Other hard theorems become more or less obvious in this context too, such as the Cayley-Hamilton Theorem: If $p$ is the characteristic polynomial for a matrix $A$, then $p(\lambda)(\lambda I-A)^{-1}$ can be easily shown to have only removable singularities, which, by the Cauchy integral formula, gives $p(A)=0$.
So why are selfadjoint and normal operators different? Selfadjoint and normal $A$ are peculiar because you can show the following: $A^{2}x=0$ iff $Ax=0$. This forces the nilpotent terms to disappear. This is a big part of the reason why the functional calculus can extend to continuous functions.