Let $(G,+,0)$ be an abelian topological ordered group, that is, $G$ is endowed with a total order $\leq$ such that, for any $a,b,c\in G$, we have that $a\leq b$ implies $a+c\leq b+c$. Moreover, $G$ is endowed with a topology such that $+:G\times G\to G$ is continuous (here we endow $G\times G$ with the product topology). Note that I am not assuming that $G$ is endowed with the order topology induced by $\leq$.
I want to know if the "absolute value" function $|\ \ |:G\to G$, given by $|a|=a$ if $a\geq0$ and $|a|=-a$ otherwise, is always continuous. For example, it is easy to see that this holds in the case of the order topology.
No, certainly not. There is no reason to expect something like this to be true, if you are not assuming that any relationship exists at all between the order and the topology.
For instance, consider $\mathbb{Q}$ with its usual order but the $p$-adic topology. Then $p^n-1$ converges to $-1$ as $n\to\infty$, but $|p^n-1|=p^n-1$ does not converge to $|-1|=1$, so the absolute value function is not continuous.
Or, as an abelian group, $\mathbb{R}$ is just a big direct sum of copies of $\mathbb{Q}$. So, we can pick some horribly discontinuous group automorphism $f:\mathbb{R}\to\mathbb{R}$. If we then define an order $\preceq$ on $\mathbb{R}$ by $x\preceq y$ iff $f(x)\leq f(y)$ (where $\leq$ is the usual order), then the absolute value function for $\preceq$ will not be continuous (it takes a bit of work to prove this; the idea is that you can find some sequence $(x_n)$ approaching a limit $x\neq 0$ such that $\preceq$ thinks the $x_n$ have the opposite sign from $x$).