How to prove the inverse of an inverse of a group element is the element itself without $a + a^{-1} = a^{-1} + a$?

I was wondering how one would go about proving the inverse of an inverse of a group element is the element itself, but without being able to use either $a + a^{-1} = a^{-1} + a$ or $0+a=a+0$?

If we only have associativity and $a + a^{-1} = 0$ (that is, if we know $a^{-1}$ is the inverse of a), I just don't see how I'd show $a = (a^{-1})^{-1}$. Any suggestions?

I'm really puzzled by this one, I have to say. I can show the equality if we have $0+a=a+0$ for all a in the group, but not without.

The claim does not follow from the assumptions you are using. For example, suppose $X$ is any set with at least two elements (one of which is called 0), and let $+$ be defined by $x+y=0$ for every $x$ and $y$, and $^{-1}$ be defined by $x^{-1}=0$ for every $x$. Then $+$ is associative, and $x+x^{-1}=0$ clearly holds for every $x$, but $(x^{-1})^{-1}=0$.

Note that we need to assume that $0 + a = a$ for all $a$ and that for all $a$ there is an $a^{-1}$ such that $a^{-1} + a = 0$ (or of course in the other order, but we cannot just assume those two conditions with the order switched in one of them).

Under these assumptions, let us check that $a + a^{-1} = 0$. We get $a^{-1} + a + a^{-1} = 0 + a^{-1} = a^{-1}$ but we can cancel $a^{-1}$ on the left dues to our assumptions, so we get $a + a^{-1} = 0$ as we wanted.

Under the hypotheses that the operation (which I will refer to as juxtaposition) is closed, associative, and satisfies the properties that there is a right identity (I will call this 1) and a right inverse is guaranteed for each element of the set $G$, it is known that $G$ is a group. We are interested in the seemingly smaller result that $a=(a^{-1})^{-1}$ for all $a\in G$.

For any $a$, we compute $$a^{-1}a=a^{-1}a1=a^{-1}a(a^{-1}(a^{-1})^{-1})=(a^{-1}1)(a^{-1})^{-1}=a^{-1}(a^{-1})^{-1}=1$$ where the associative law was used to cancel middle terms. This proves that the inverse $a^{-1}$ applies to both sides of a. In particular, $a^{-1}(a^{-1})^{-1}=1=(a^{-1})^{-1}a^{-1}$, so we can now prove the desired statement.$$a=a1=a(a^{-1}a)=1a=((a^{-1})^{-1}a^{-1})a=(a^{-1})^{-1}$$

Notice that in doing so, we prove that $a1=a=1a$; all we did was show that $G$ is a group and do the usual proof of the fact that the inverse of the inverse recovers the original element.