Your formula is based on the identity for $s\in\mathbb N$,
$$
\partial^s=(1+(-1)+\partial)^s=\sum_{m\ge 0}C_s^m((-1)+\partial)^m=\sum_{m\ge 0\atop 0\le k\le m}C_s^mC_m^k(-1)^{m-k}\partial^k.
\tag{$\flat$}$$
It is interesting to compare it to the Fourier multiplier method
$$
\partial =(2iπ\xi)^s=(2π)^s\exp{(s\text{Log}(i\xi))}
$$
with the principal determination of the logarithm ($i\xi\in i\mathbb R$). We have
$$
\widehat{\partial^s u}=(2iπ\xi)^s\hat u(\xi),\quad
(\partial^s u)(x)=\int e^{2iπ x\xi}(2iπ\xi)^s\hat u(\xi)d\xi.
\tag{$\sharp$}
$$
Of course, using here the notation $D=\frac{\partial}{2iπ\partial x}$,
we can also write as you did
$$
\partial^s=(1+(-1)+2iπ D)^s
$$
but the radius of convergence of $(1+z)^s$ is 1 and you will be forced to drastic assumptions on the decay of the derivatives to have something convergent.
Note however that the coefficients in the expansion are the same as in $(\flat)$.
A very natural requirement is that $\partial^s$ sends the homogeneous Sobolev space $\dot H^m$
into
$\dot H^{m-s}$: we can define in $n$ dimensions for $m>-n/2$
$$\dot H^m=\{u\in\mathscr S'(\mathbb R^n),\hat u\in L^2_{loc}, \vert\xi\vert ^m \hat u\in L^2(\mathbb R^n)\}.
$$
In one dimension, for $s\in\mathbb R$, $m>s+\frac12$, Definition $(\sharp)$ provides the continuity of $\partial^s$ from $\dot H^m$
into
$\dot H^{m-s}$. I believe that with Formula $(\flat)$, since the homogeneity is not obvious, it should not be so easy to prove that continuity property. On the other hand, for $s\in \mathbb R$, the property
$$
\partial ^s\partial^{-s} u=u
$$
is obvious with $(\sharp),$
provided we assume that $u$ belongs to $\dot H^{m}$ with $m>\frac12+\vert s\vert$,
which is a mild requirement compared to the decay estimates on derivatives that you will need to have $(\flat)$ convergent.
