Hint $\ $ The vector space arises by transport of structure from the vector space $\,(0,\infty)\,$ to $\,(8,\infty)\,$ along the shift-bijection $\,s(v)= v+8,\,$ with inverse $\,t(v) = v-8.\,$ The former vector space has the simpler vector operations given by $\ a u \oplus b v = u^a v^b.\,$ Transporting is defined so that the the maps $\,s,t\,$ are vector space isomorphisms, e.g. $\ t(u\boxplus v) = t(u)\oplus t(b)\,$ so the transported addition is
$$ u\boxplus v\, =\, t^{-1}(t(u)\oplus t(v))\, =\, (u\!-\!8)(v\!-\!8)+8\, =\, uv -8(u\!+\!v)+ 72$$
Thus you can compute inverses straightforwardly by first performing the simpler inverse computation in the space $\,(0,\infty)\,$ then shift-transporting the solution to $\,(8,\infty),\,$ as above.
Remark $\ $ Although this problem is so simple that one could skip the conceptual transportation viewpoint and do the computation by brute-force, this would be much more difficult to do in more complex examples. For example, in Gauss's Disq. Arith. the proof of associativity of composition of binary quadratic forms comprises many pages of unilluminating abstruse calculations, whereas nowadays this can be done by simply by transporting the class group structure from ideals to primitive binary quadratic forms.
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.
Best Answer
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.