Does the First Functor Law follow from the Second?

No

data Break a = Yes | No

instance Functor Break where
   fmap f _ = No

clearly the second law holds

   fmap (f . g) = const No = const No . fmap g = fmap f . fmap g

but, the first law does not. The problem with your argument is not all x' are of the form fmap f x

No, it only works in one direction.

Consider this Functor instance:

data Foo a = Foo Int a

instance Functor Foo where
    fmap f (Foo _ x) = Foo 5 (f x)

It satisfies the second law but not the first one.

The last step in your proof is invalid -- you showed that fmap id x' = x', but this is restricted to x's that are returned from fmap in the first place, not arbitrary values.