Consider the two structures $(\mathbb{R};f)$ and $(\mathbb{R};-)$, where $f$ represents additive inverse, and $-$ represents binary subtraction. Certainly, $f$ can be defined from $-$. Can $-$ be defined by $f$? I don't think it can, but I want a formal proof that no possible definition works.
Edit: It is very clear to me that $-$ can't be expressed by a term in $f$, but I am wondering if the graph of $-$ can be defined by a first-order formula in $f$.
To move this off the unanswered queue, I'll expand Eric Wofsey's comment into an answer:
Definable structure is always preserved by automorphisms. To show that subtraction can't be defined from inversion alone, we just need to find some permutation of $\mathbb{R}$ respecting inversion but not respecting subtraction. One example which I quite like is the function $$b(x)=\begin{cases} x & \mbox{ if } x\in\mathbb{Q},\\ f(x) & \mbox{ if } x\not\in\mathbb{Q}.\\ \end{cases}$$ To see that $b$ does not in fact preserve subtraction just note that e.g. $b(\pi-(\pi-1))=b(1)=1$ but $b(\pi)-b(\pi-1)=-\pi-(1-\pi)-1$.
There are two quick notes worth making at this point:
First, the above analysis has very little to do with first-order logic. Isomorphism invariance of satisfaction, and consequently automorphism invariance of definability, is (almost always) one of the defining criteria for something to be a logic in the first place. So really what the above shows is that inversion can't define subtraction in any logical system.
Second, note that inversion is really combinatorial rather than algebraic when taken on its own: the structure $(\mathbb{R};f)$ is better thought of as just a set equipped with an equivalence relation, namely $\vert x\vert=\vert y\vert$, which partitions it into continuum-many pairs and one singleton. Thought of this way it's clear that we shouldn't expect any expressive power (and moreover this motivates non-interpretability, rather than mere non-definability, results).