Suppose we have a field with a total order which verifies 


If $x,y\geq 0$, then $x+y\geq 0$ (here we relax the compatibility property for the addition).


If $x,y\geq 0$, then $xy\geq 0$.


Suppose also that every upper bounded nonempty set has a supremum.


We don't assume that $x\leq 0, -x\leq 0$ implies $x=0$.


Is this field the ordered field of real numbers?



Axioms for the real numbers
The answer is no. E.g., let $R$ be the field of all real numbers with the strict total order $\prec$ defined as follows: for any $x$ and $y$ in $R$ such that $x<y$ (where $<$ is the usual order on $R$), let $x\prec y$ if $y\ge0$, and $y\prec x$ if $y<0$. Then all your conditions on the total order will hold, but $(R,\prec)$ will not be an ordered field. 


**Details:**
We see that the order $\prec$ preserves the usual order among the nonnegative real numbers, but reverses the usual order among the negative real numbers, also keeping the negative numbers less than the nonnegative ones. 


Since the $\prec$-nonnegative numbers are the same as the $<$-nonnegative ones, your first two conditions, concerning the addition and multiplication operations, hold. 


Now take any nonempty $\prec$-upper-bounded subset $A$ of $R$. If there is a nonnegative number in $A$, then $A$ is $<$-upper-bounded, and $\sup\_{\prec} A=\sup\_{<} A\in R$; the subscripts $\_\prec$ and $\_<$ indicate here the order with respect to which the supremum or infimum is taken. 


Otherwise, that is, if there is no nonnegative number in $A$, then $A$ is always $\prec$-upper-bounded by $0$. If now $c:=\inf\_< A\ne-\infty$, then $c=\sup\_\prec A\in R$. Finally, if $c=-\infty$, then $\sup\_\prec A=0\in R$. 


So, any nonempty $\prec$-upper-bounded subset $A$ of $R$ has $\sup\_\prec A\in R$. Thus, your third, and final, condition on the total order holds as well. 


However, $(R,\prec)$ is not an ordered field, because $-1\prec-2$ but $-1+1\not\prec-2+1$. 


