Let E be the forward shift operator on $x$ defined by $Ef(x) = f(x+1)$. Similarly, let $\delta$ be the forward difference operator such that $\delta f(x) = f(x+1) - f(x)$ and the inverse operator $\sum$ be the operator that undoes the difference operation. Prove that $\sum(f(x)\delta g(x))$ = f(x)g(x) - $\sum(Eg(x)\delta f(x))$.
Am I correct in assuming that this is a sort of integration by parts rule for these operators?
Yes, that’s exactly what it is. And you derive it in the same way, as a consequence of the product rule for the difference operator:
$$\begin{align*} \Delta\big(f(x)g(x)\big)&=f(x+1)g(x+1)-f(x)g(x)\\ &=f(x+1)g(x+1)-f(x)g(x+1)+f(x)g(x+1)-f(x)g(x)\\ &=f(x)\Delta g(x)+Eg(x)\Delta f(x)\;. \end{align*}$$