(Since I usually use left-orderings, the following answer uses left-orderings)

No, when $<$ is a left-ordering which is not a bi-ordering then you can always find elements $x,y,z$ with
$x <z $ and $y<z$ such that $xy>z^2$.

To see this, since $<$ is not a bi-ordering we can always find an element $a,b$ such that
$1<a$ but $b^{-1}ab <1$ (so $ab < b < a^{-1}b$).
Now $a^{-2} < 1$, $ab < b$ but $(a^{-2})(ab) = a^{-1}b > b$.

On the other hand, it should be mentioned that somewhat related property holds for Conradian orderings;
if < is a Conradian left-ordering, then $b < ab^{2}$ holds for every $1<a,1<b$. This is stated and proved in Proposition 3.7 of [On the dynamics of (left) orderable groups][1] 


  [1]: http://www.numdam.org/item/AIF_2010__60_5_1685_0/