TITLE: Does there exist an operation that could turn the set of all negative real numbers into an abelian group?
QUESTION [11 upvotes]: The answer is no for familiar operations of addition and multiplication. But could there exist any other operation that could turn the set of all negative real numbers into an abelian group. If yes, what is it? If no, how could I prove it?

REPLY [6 votes]: I add this "answer" (too long for a comment) just to make some more explicit the result used in the nice answers above.
Claim. Let $(G,\cdot)$ be a group and $\tilde G$ a set. If there is a bijection $f\colon G\to \tilde G$, then the operation:
$$\tilde g*\tilde h:=f(f^{-1}(\tilde g)\cdot f^{-1}(\tilde h)), \space\space\space\forall \tilde g,\tilde h \in \tilde G\tag 1$$

defines a group structure in $\tilde G$;
$(\tilde G,*) \cong (G,\cdot)$.

Proof.

Closure: by definition $(1)$.

Associativity:


\begin{alignat}{1}
(\tilde g*\tilde h)*\tilde k &=f\color{red}{(}f^{-1}(\tilde g*\tilde h)\cdot f^{-1}(\tilde k)\color{red}{)} \\
&=f\color{red}{(}f^{-1}\color{blue}{(}f\color{cyan}{(}f^{-1}(\tilde g)\cdot f^{-1}(\tilde h)\color{cyan}{)}\color{blue}{)}\cdot f^{-1}(\tilde k)\color{red}{)} \\
&=f\color{red}{(}f^{-1}(\tilde g)\cdot f^{-1}(\tilde h)\cdot f^{-1}(\tilde k)\color{red}{)} \\
&=f\color{red}{(}f^{-1}(\tilde g)\cdot \color{blue}{(}f^{-1}(\tilde h)\cdot f^{-1}(\tilde k)\color{blue}{)}\color{red}{)} \\
&=f\color{red}{(}f^{-1}(\tilde g)\cdot (f^{-1}f)\color{blue}{(}f^{-1}(\tilde h)\cdot f^{-1}(\tilde k)\color{blue}{)}\color{red}{)} \\
&=f\color{red}{(}f^{-1}(\tilde g)\cdot f^{-1}\color{blue}{(}f\color{cyan}{(}f^{-1}(\tilde h)\cdot f^{-1}(\tilde k)\color{cyan}{)}\color{blue}{)}\color{red}{)} \\
&=f\color{red}{(}f^{-1}(\tilde g)\cdot f^{-1}\color{blue}{(}\tilde h*\tilde k\color{blue}{)}\color{red}{)} \\
&=\tilde g*(\tilde h *\tilde k), \space\space\space\forall \tilde g,\tilde h,\tilde k\in \tilde G \\
\tag 2
\end{alignat}

Unit: let's define $e_{\tilde G}:= f(e_G)$; then:

\begin{alignat}{1}
\tilde g * e_{\tilde G} &= f(f^{-1}(\tilde g)\cdot f^{-1}(e_{\tilde G})) \\
&= f(f^{-1}(\tilde g)\cdot f^{-1}(f(e_G))) \\
&= f(f^{-1}(\tilde g)\cdot e_G) \\
&= f(f^{-1}(\tilde g)) \\
&= \tilde g, \space\forall \tilde g \in \tilde G \\
\tag 3
\end{alignat}
and $e_{\tilde G}$ nicely behaves as unit.

Inverse: $\forall \tilde g \in \tilde G$, let's define $\tilde g^{-1}:=f((f^{-1}(\tilde g))^{-1})$; then:

\begin{alignat}{1}
\tilde g*\tilde g^{-1} &= f(f^{-1}(\tilde g)\cdot f^{-1}\color{red}{(}f((f^{-1}(\tilde g))^{-1})\color{red}{)}) \\
&= f(f^{-1}(\tilde g)\cdot (f^{-1}(\tilde g))^{-1}) \\
&= f(e_G) \\
&= e_{\tilde G}, \space\forall \tilde g \in \tilde G \\
\tag 4
\end{alignat}
and $\tilde g^{-1}$ nicely behaves as inverse of $\tilde g$.
Therefore, $(\tilde G, *)$ is a group. Finally, $\psi:=f^{-1}$ is a group homomorphism, because (by $(1)$) $\psi(\tilde g*\tilde h)=\psi(\tilde g)\cdot \psi(\tilde h), \space\forall \tilde g,\tilde h\in \tilde G$, and thence $(\tilde G,*)\cong (G,\cdot)$.

Example
$(G,\cdot)=(\mathbb{R},+)$, $\space\space\tilde G=\mathbb{R}_{<0}$, $\space\space v=f(x)=-\operatorname{exp}(x)$. Then:

Group operation (see $(1)$):
\begin{alignat}{1}
v*w &= -\operatorname{exp}(x+y) \\
&= -\operatorname{exp}(x)\operatorname{exp}(y) \\
&= -vw
\end{alignat}

Unit:


$$0_{(\mathbb{R}_{<0},*)}=-\operatorname{exp}(0)=-1$$

Inverse:

$$v^{-1}=-\operatorname{exp}(-x)$$
By the claim:
$$(\mathbb{R}_{<0},*,-1)\cong (\mathbb{R},+,0)$$