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.
For the zero vector: you can determine it from any of the two equations (which would make you check that they are compatible, which they are).
If $(a_1,a_2)+(z_1,z_2)=(a_1,a_2)$, this becomes $$(a_1+z_1,a_2z_2)=(a_1,a_2).$$ So you have $a_1+z_1=a_1$, $a_2z_2=a_2$. As you can do this for any choice of $a_1,a_2$, it follows that $(z_1,z_2)=(0,1)$.
Or, easier: $$(z_1,z_2)=0\cdot(1,1)=(0,1^0)=(0,1).$$
For additive inverse: if $(a_1,a_2)+(b_1,b_2)=(0,1)$, we have
$$
(a_1+b_1,a_2b_2)=(0,1),
$$
that is $a_1+b_1=0$, $a_2b_2=1$. So $b_1=-a_1$, $b_2=1/a_2$. So for $(a_1,a_2)$ with $a_2\ne0$, its additive inverse is $$\left(-a_1,\frac1{a_2}\right).$$
Best Answer
In standard vector spaces you have only addition and scalar multiplication, so the only inverse is the additive inverse.
$$ \mathbf{v}+(-\mathbf{v})=\vec{0} $$
However, in geometric algebra vectors exist as a subset of a larger set of objects including scalars and "multi-vectors" in which a product is defined. This product subsumes the scalar product of standard vector theory.
In this context, some non-zero vectors have multiplicative inverses.