These are the axioms that I'm familiar with for vector spaces:
this is my problem:
So this IS closed under additionright?
If u and v are vectors (u could be (x,y) where x and y are both $\geq 0$), then if we add them together, then they are both $\geq 0$ right?
But it may not be closed under multiplication right? Like if we multiply (0,4) by a scalar -1, then the resulting vector is not in the set anymore right?
Yes, the scalar multiplication is not closed in the set $V = \{ (x,y) \, : \, x \geq 0, y \geq 0 \}$ and your example works perfectly.
Another axiom that does not hold is the existence of an additive inverse, for example consider the vector $(1,2)$ there is no vector $(a,b)$ in $V$ such that $(1,2) + (a,b) = (0,0)$, the only vector with this property (prove it) is $(-1,-2)$ but it is not in $V$.