I have an operation $\boxdot$ over $\mathbb R$ which makes $(\mathbb R,<,+,\boxdot)$ into an ordered division ring (field without commutativity for $\boxdot$, or rather such commutativity is not proven yet). Of course, $\boxdot$ is compatible with the order on $\mathbb R$.
May I conclude that $\boxdot$ is just the ordinary multiplication, so that I get the field $(\mathbb R,<,+,\cdot)$?
Every ordered ring with underlying additive group $(\Bbb R,+)$ and the standard ordering has multiplication of the form $a\boxdot b= \lambda ab$ for a unique $\lambda \geq 0$ and is hence commutative.
Lemma: Order-preserving endomorphisms of $(\Bbb R,+,<)$ are linear
Let $\varphi$ be any order-preserving group endomorphism of $(\Bbb R,+)$. We can exclude the case that $\varphi \equiv 0$. I claim that there exists a unique $\lambda \in \Bbb R_{>0}$ such that $\varphi(x)=\lambda x$ for all $x \in \Bbb R$. The argument is quite similar to the proof that the identity is the only field endomorphism of $\Bbb R$. Uniqueness is clear as $\lambda=\varphi(1)$ is necessary. We may wlog assume that $\varphi(1)=1$ (else replace $\varphi$ by $\frac{1}{\lambda} \varphi$).
If there is any $x \in \Bbb R$ such that $\varphi(x) \neq x$, then either $\varphi(x) >x$ or $\varphi(x)<x$. I'll treat the first case, the other being analogous. Choose $q \in \Bbb Q$ such that $\varphi(x)>q>x$. From $\varphi(1)=1$ it follows $\varphi(q)=q$. But because $\varphi$ is order-preserving, we obtain from $q>x$ the inequality $\varphi(q)>\varphi(x)$. Putting these together yields the contradiction $q=\varphi(q)>\varphi(x)>q$.
Using the lemma, we can finish the argument and actually prove something stronger than just a statement about division rings. If $(\mathbb R,<,+,\boxdot)$ is an ordered ring, not necessarily commutative, not necessarily unital. Then for any $r \in \Bbb R_{>0}$, the map $x \mapsto r \boxdot x$ is an endomophism of the ordered group $(\Bbb R,+,<)$. Thus there exists a unique $\lambda_r \in \Bbb R_{>0}$ such that $r \boxdot x=\lambda_r x$ for all $x$.
If we fix $r,s \in \Bbb R_{>0}$, we obtain for any $x \in \Bbb R$:
$(s \boxdot r) \boxdot x= s \boxdot (r \boxdot x)=s \boxdot (\lambda_r x)=\lambda_s \lambda_r x$. We extend the map $r \mapsto \lambda_r$ to all reals by setting $\lambda_{-r}=-\lambda_r$ if $r<0$ and $\lambda_0=0$. Let's denote the map $\Bbb R \to \Bbb R, r \mapsto \lambda_r$ by $\psi$. Then we have shown above that $\psi(s \boxdot r)=\psi(s) \psi(r)$ for all $r,s$. But if we forget about multiplications, $\psi$ is an order-preserving endomorphism of $(\Bbb R,+)$. To see this note that for $s,r \in \Bbb R$, we have for any $x \in \Bbb R$: $(r+s)\boxdot x = r\boxdot x + s \boxdot x = \lambda_r x+\lambda_s x=(\lambda_r+\lambda_s)x$ and thus $\psi(r+s)=\lambda_{r+s}=\lambda_r + \lambda_s = \psi(r)+\psi(s)$.
Thus $\psi(x)=\lambda x$ for $\lambda:=\psi(1)>0$ (omitting the trivial case $\psi(1)=0$). We obtain from $\psi(s \boxdot r)=\psi(s) \psi(r)$ the conclusion that $s \boxdot r=\lambda (sr)$, which answers the question raised in a comment by Griboulis in the affirmative.
I believe the answer is yes and that Gribouillis comment is correct. This follows from a more general fact that any Archimedean totally ordered division ring is commutative. I will try to sketch a proof.
Note that $(\mathbb{Z},+)$ has only one notion of multiplication that makes it a ring with $1$ being the unit. In particular this means there can be only one operation which makes $(\mathbb{Z},+,<)$ and ordered ring. We can extend this to show that if $\boxdot$ is an operation that makes $(\mathbb{Q},+,<,\boxdot)$ a division ring then it must be of the form $a\boxdot b=q\cdot a\cdot b$ where $q\in\mathbb{Q}$ is non zero. Finally, let $u\in \mathbb{R}$ denote the multiplicative identity with respect to $\boxdot$. You have $a\boxdot b=u^{-1}\cdot a\cdot b$ for all $a,b\in u\mathbb{Q}=\{u\cdot q: q\in \mathbb{Q}\}$ since $u\mathbb{Q}$ is isomorphic as an ordered abelian group to $ \mathbb{Q}$. For every $\epsilon>0$ we can find $q\in \mathbb{Q}$ such that: $$(u\cdot q)\boxdot (u \cdot q)=u^\cdot q^2<\epsilon$$ By the properties of ordered rings we have: $$\forall s,t\in (-uq,uq)_{\mathbb{R}}\,(|s\boxdot t|<\epsilon )$$ It follows that $\boxdot $ is continuous at $(0,0)$ which implies it is continuous on all of $\mathbb{R}$ (use the translation property of ordered rings). By density of $u\mathbb{Q}$ in $\mathbb{R}$ you have that for all $s,t\in \mathbb{R}$ we have $s\boxdot t=u^{-1}\cdot s \cdot t$
