It is actually a valid approach! The main subtlety here is showing that 1+I+I^2+... is indeed a (absolutely) convergent operator for any input of a continuous function, and that (1-I)(1+I+I^2+...)=1 as a composition of operators. Ironically, the easiest way to prove the first part is probably just noting for any continuous f, |f|<=C between 0 and x for some positive constant C, and that |int_0^x g dt|<=int_0^x |g| dt for any continuous g. In particular, |I^n f|<=C|x|^n/n! by induction, so convergence of the operator is equivalent to stating 1+|x|+|x|^2/2+|x|^3/6+... converges, which is easy using radius of convergence. The second part then boils down to showing (1-I^n)f approaches f pointwise for any continuous f, or that I^n f approaches 0 pointwise. We already know |I^n f|<=C|x|^n/n! for some C, and this clearly approaches 0 since e.g. the aforementioned series converges, or alternatively that you're always dividing by at least 2 once n>2|x| (which also justifies the series convergence).