Does there exist an operation that could turn the set of all negative real numbers into an abelian group? 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?
 A: 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:
$$(\mathbb{R}_{<0},*,-1)\cong (\mathbb{R},+,0)$$
A: If you genuinely allow any operation, then the answer is yes for a silly (but important!) reason: we can lift structure along bijections. Specifically, fix some bijection $f:\mathbb{R}\rightarrow\mathbb{R}_{<0}$. Then we can define an addition map $\oplus$ as follows: $$a\oplus b=f(f^{-1}(a)+f^{-1}(b)).$$ The map $f$ shows that "$\mathbb{R}$ with $+$ looks identical to $\mathbb{R}_{<0}$ with $\oplus$" - or, in more precise language, the two structures $(\mathbb{R};+)$ and $(\mathbb{R}_{<0};\oplus)$ are isomorphic. A fortiori they have the same general algebraic properties: in particular, $(\mathbb{R}_{<0};\oplus)$ is an abelian group since $(\mathbb{R};+)$ is.

*

*A notational comment: the expression "$(A; [\DeclareMathOperator{\stuff}{stuff}\stuff])$" indicates that $A$ is the underlying set of the structure involved and $[\stuff]$ is the list of operations and relations on that set, with different things in $[\stuff]$ being separated by commas (in contrast with the semicolon separating $A$ and $[\stuff]$). So, for example, "$\mathbb{R}$ as an ordered ring" would be written as "$(\mathbb{R};+,\cdot,<)$."

Basically, when we ask "Does such-and-such structure exist on the set $X$?," all that really matters is the cardinality of $X$: whenever $X_1,X_2$ are in bijection with each other, the answer for $X=X_1$ will be the same as the answer for $X=X_2$.
Things get more interesting if we ask for the desired structure to satisfy some additional properties. For example, we might want the relevant operations - the (binary) group operation $\oplus$ and the corresponding (unary) inverse operation - to be continuous with respect to the usual topology on $\mathbb{R}_{<0}$. Now it's not the case that any old bijection $\mathbb{R}\rightarrow\mathbb{R}_{<0}$ will do, since a really messy bijection might turn the continuous $+$ into something highly discontinuous; we need to be a bit more careful. As a matter of fact, however, we can find one which does the job (consider the map $f(x)=-e^x$).
A: A concrete example of an abelian group structure on negative reals is
$$x\circ y:=-xy$$
A: If you ask whether the negative reals form a vector space over the rationals without specifying the operations, the obvious way to view the question is using the standard operations of addition and multiplication.  You point out correctly that this fails.  If you are allowed to define new operations, all that is left is the cardinality of the negative reals, which is $\mathfrak c$ like all the reals.  We know that the standard reals form a vector space over the rationals, so we just need to define a bijection $f: \Bbb{R^- \leftrightarrow R}$.  Then we define $\oplus$ on the negative reals as $x \oplus y=f^{-1}(f(x)+f(y))$ and $\otimes$ as $x \otimes y=f^{-1}(f(x)\cdot f(y))$  These operations, with $f^{-1}(0)$ as the identity for $\oplus$ and $f^{-1}(1)$ as the identity for $\otimes$ will make the negative reals a vector space over the rationals.  I leave defining scalar multiplication and inverses to you.
