Simplifying to Linear Differential Operator?
It seems that $\;\phi^{-1} = 1/\phi\;$ and that the product operator (i.e function)
happens to be at the same time a solution. Then everything on that page seems indeed to be correct and:
$$
e^{c\phi E \phi^{-1}}f(x,y)= \left[ \sum_{n=0}^\infty \left(\phi\, cE \,\phi^{-1}\right)^n / n! \right] f(x,y)
$$
Where:
$$
\left(\phi\,cE\,\phi^{-1}\right)^n = \left(\phi\,cE\,\phi^{-1}\right)
\left(\phi \,cE\,\phi^{-1}\right)\left(\phi\,cE\,\phi^{-1}\right)\cdots =\\
\phi\,cE\cdot 1/\phi\ \cdot\phi\,cE\cdot 1/\phi \cdot \phi\,cE\cdot 1/\phi\cdots = \phi\left(cE\right)^n\phi^{-1}
$$
Because $\,1/\phi\cdot \phi = 1$ . Hence the formula follows:
$$
e^{c\phi E \phi^{-1}}f(x,y)=
\left[ \sum_{n=0}^\infty \phi\left(cE\right)^n \,\phi^{-1}/n! \right]f(x,y) =
\phi \left[ \sum_{n=0}^\infty \left(cE\right)^n /n! \right] \phi^{-1} f(x,y) \\ =
\phi (x,y) e^{cE}[\phi^{-1}(x,y)f(x,y)]
$$