This vector space is nice, never encountered such an example.
If you found that the zero vector is $-2$, the additive inverse is not too far away. You actually don't need to care about scalar multiplication for this part.
The problem of finding the additive inverse is to find a function $i:V\to V$ such that for every $u\in V$, you have $u\boxplus i(u)=e$, where $e$ is the zero vector of $(V,\boxplus,\boxdot)$.
So the first step is to find the zero vector, which you claim you had. Just to be sure, I show here how it is found: for all $u\in V$, we want $u\boxplus e=u$. This means for all $u\in \mathbb R$, we have $u+e+2=u$. We immediatly get $e=-2$, by choosing any $u$ (for instance it suffices to look at $u=0$).
Now we want to find the function $i$ which gives us the additive inverse of any vector of $V$.
Given $u$, we need to find $i(u)$ such that $u\boxplus i(u)=e$. This last equation translates $u+i(u)+2=-2$. we solve the equation for $i(u)$, and we get $i(u)=-4-u$, for all $u\in V$. So this shows that the additive inverse in $V$ is given by the formula $i(u)=-4-u$. Notice that for instance $i(e)=-4-(-2)=-4+2=-2=e$, which is normal: zero is always its own inverse.
You're right that the way to solve the first one is to check the axioms individually. Make a couple of third degree polynomials $u(x) = ax^3 + bx^2 + cx + d$ and $v(x) = ex^3 + fx^2 + gx + h$ and play around with them.
What is $u(x) + v(x)$? Is it a third degree polynomial? If so, you have closure under addition.
What is the additive identity? Is that a third degree polynomial?
What do you get when you multiply $u(x)$ by a scalar $k$? Is that a third degree polynomial?
Et cetera.
Best Answer
Closure under addition and scalar multiplication is enough, since the additive inverse and the zero vector may be obtained by multiplying vectors with the scalars $-1$ and $0$, respectively. You do have to make sure the set is not empty, though, and the easiest vector to check whether it's actually in there is usually the zero vector.