I want to show that for every $\lambda \neq 0$ and $g\in L^{2}(\mathbb{R})$, there is a $f \in L^{2}(\mathbb{R})$ such that
$$e^{-|x+1|}f(x) - \lambda f(x+1) = g(x)$$
I tried looking at the operator for which this is the resolvent and using the adjoint I could show it has dense range, but I couldn't show it was closed as well. Can one directly find a function $f$ satisfying this functional equation?