> Isn't (1-ar(L)) == (1 - ar_1 ... - ar_p)? 
only if applied to a constant

(1-ar(L)) x_t =  1 * x_t - ar_1 * x_{t-1} - ... - ar_p * x_{t-p}
if x_t = const for all t, then
(1-ar(L)) x_t == (1 - ar_1 ... - ar_p) * const

short hand notation with 1 instead of lag operator L
(1-ar(1)) == (1 - ar_1 ... - ar_p)