Better would be the following:Am I understanding it correct when I do 1(x, y, z) = (x, y, z) != (0, 0, 0).

According to the definition of your candidate space,

It does not matter what x, y, or z are. This causes axiom 10 to fail.

Now, axiom 7 holds, actually. Everything on both sides there will be the zero vector. The same is true of axioms 8 and 9.