One standard definition of a linearly ordered group is that the order $\leq$ obeys the law
(1) $a\leq b\implies ac\leq bc$.
Suppose (1) only holds for positive c. Must it also hold for negative c?
I've been able to show that (1) holds if
(2) $a\leq 1\implies a^{-1}\geq 1$
but I can't prove (2) knowing (1) only for positive c. However, I can't think of a counter-example, and it seems weird to have (2) not be true.
On
Let $G=\{0,1\}$ under addition modulo 2 (So $G=\mathbb{Z}_2$). Let $\leq$ be defind by $1\leq 0$, $1\leq 1$, $0\leq 0$. This is a perfectly good linear ordering on the set $G$.
This satisfies your condition (1) for all positive $c$ because $0$ is the only positive element in this group and it is the identity. $1$ is its own inverse and so $1^{-1}\leq 0$. It follows that your condition (2) does not follow from condition (1) for all positive $c$.
Taken to the extreme, we can just make an arbitrary ordering of any non-trivial group $G$ as long as $g\leq e$ for all $g\in G$ where $e$ is the identity of $G$. Such an ordering will trivially satisfy your condition (1) for all positive elements $c$ and will never saitisfy your condition (2).
Consider the multiplication group $(\mathbb{R}-\{0\},\times)$, the order is the usual one. Then $a<b$ implies $ac<bc$ for $c>1$ but $ac>bc$ for $c<0$ (note the meaning of negative is now <1).