This is a fuller explanation of what axblount discovered empirically.
You have a positive integer $n=\sum_{k=0}^ma_k10^k$, where $a_m\ne 0$. Think of the digits as elements of a sequence $\langle a_0,\dots,a_m\rangle$. Your basic operation is applying the forward difference operator and changing the sign.
Your first derived sequence is therefore $\langle-\Delta a_0,\dots,\Delta a_{m-1}\rangle$.
Now in general we have $\Delta(-x_k)=-x_{k+1}-(-x_k)=x_k-x_{k+1}=-\Delta x_k$, so $-\Delta(-x_k)=\Delta x_k$, and your second derived sequence is $\langle\Delta^2 a_0,\dots,\Delta^2a_{m-2}\rangle$. Your third is $\langle -\Delta^3a_0,\dots,-\Delta^3a_{m-3}\rangle$, and in general the $n$-th is $\big\langle(-1)^n\Delta^na_0,\dots,(-1)^n\Delta^na_{m-n}\big\rangle$. In particular, your final single value, before taking the digital root, is $(-1)^m\Delta^ma_0$.
Now it’s well-known (and easily proved by induction) that in general
$\Delta^n x_k=\sum_{i=0}^n(-1)^i\binom{n}ix_{k+n-i}\;,$
so
$\begin{align*} (-1)^m\Delta^ma_0&=(-1)^m\sum_{i=0}^m(-1)^i\binom{m}ia_{m-i}\\ &=(-1)^m\sum_{i=0}^m(-1)^{m-i}\binom{m}ia_i\\ &=\sum_{i=0}^m(-1)^i\binom{m}ia_i\;. \end{align*}$
This is the number whose digital root you take in the final step.