(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
[Added] More concrete answer
To avoid confusion, from now on, let us use right ordering (as opposed to the original answer)
Let $(G,<)$ be a right-ordered group. Assume that $<$ is not a bi-ordering, so there exist $s,t \in G$ such that $1<s$ but $t>ts$. Then
$\cdots<ts<t<ts^{-1}<ts^{-2}<\cdots$.
So by multiplying $t$ from the right we get
$\cdots<tst<t^2<ts^{-1}t<ts^{-2}t<\cdots$.
Now put $x=ts$, $y=s^{-2}t$, $z=t$. Then
$x = ts < t=z$
$y=s^{-2}t<t=z$ (because $s^{-2}<1<s$ and $<$ is a right-ordering), and
$xy=(ts)(s^{-2}t)=ts^{-1}t>t^{2}=z^2$.
This gives a negative answer to the original question; $x<z,y<z$ but $xy>z^{2}$ (note that Conradian property is not needed).
To get a negative answer to the refined question (elements $x,y,z$ with $1<x<y<z$ with $xy>z^{2}$), note that in the above construction, we can take $1<t$ so that $1<x=ts$ holds. Since $x<y \iff ts<s^{-2}t \iff tst^{-1} < s^{-2}$, by taking $G$ so that $1<s,t$ and $tst^{-1}<s^{-2}<1$ holds, we get $1<x<y<z$ with $xy>z^{2}$.
(For example, let $G=\langle s,t \: | \: s \in \mathbb{Q}, t \in \mathbb{Z}, tst^{-1}=s^{-3} \rangle$ with natural extension right order.)