well you have to show closure, that is if f(x),g(x) is in P1, so is f(x)+g(x).

now (a_{0}+ a_{1}x) + (b_{0}+ b_{1}x) = (a_{0}+b_{0}) + (a_{1}+b_{1})x, no matter if P1 is closed or not.

however, you are correct that if a_{0}- 2a_{1}= b_{0}- 2b_{1}= 0, then:

(a_{0}+b_{0}) - 2(a_{1}+b_{1}) = (a_{0}-2a_{1}) + (b_{0}-2b_{1}) = 0 + 0 = 0.

if you have already established that V = {a_{0}+ a_{1}x : a_{0},a_{1}in F} (you don't SAY what F is, but presumably it is the field of real numbers, it doesn't HAVE to be, though...there ARE other fields) is a vector space, you don't have to "re-prove" commutativity of vector addition or associativity of vector addition, since these things hold for ALL elements of V (including those that happen to be in P1). we say such properties are "inherited" by any subset.

you WILL have to show that P1 has a 0-vector (or at the very least that at least ONE polynomial of degree 1 or less, or the 0-polynomial, satisfies your membership requirement for P1).

if you show "closure under scalar multiplication", then the two distributive laws, and the unit scalar multiplication law are also "inherited" properties.