You misunderstand how boolean expressions work; they don’t work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:
if x == 1 or y == 1 or z == 1:
x and y are otherwise evaluated on their own (False if 0, True otherwise).
You can shorten that using a containment test against a tuple:
if 1 in (x, y, z):
or better still:
if 1 in {x, y, z}:
using a set to take advantage of the constant-cost membership test (i.e. in takes a fixed amount of time whatever the left-hand operand is).
Explanation
When you use or, python sees each side of the operator as separate expressions. The expression x or y == 1 is treated as first a boolean test for x, then if that is False, the expression y == 1 is tested.
This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first.
However, even if this were not the case, and the expression x or y or z == 1 was actually interpreted as (x or y or z) == 1 instead, this would still not do what you expect it to do.
x or y or z would evaluate to the first argument that is ‘truthy’, e.g. not False, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).
So for the values x = 2; y = 1; z = 0, x or y or z would resolve to 2, because that is the first true-like value in the arguments. Then 2 == 1 would be False, even though y == 1 would be True.
The same would apply to the inverse; testing multiple values against a single variable; x == 1 or 2 or 3 would fail for the same reasons. Use x == 1 or x == 2 or x == 3 or x in {1, 2, 3}.