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Let $[[(a,b)]]_{\ominus}$, $[[(c,d)]]_{\ominus}$ and $[[(e,f)]]_{\ominus}$ be formally constructed rational numbers - equivalence classes of ordered pairs under the $(z_1,z_2)\,{\ominus}\,(z_3,z_4){\iff}z_1*z_4=z_2*z_3$ relation, where $z_1,z_3\,{\in}\,\mathbb{Z}$ and $z_2,z_4\,{\in}\,\mathbb{Z}/{\{0\}}$. For convenience, I'll use $[a,b]$, $[c,d]$ and $[e,f]$ to mean those equivalence classes.

Let $(\mathbb{Q},\mathbb{Q},{\le})$ be a relation defined by $[m_1,n_1]\,{\le}\,[m_2,n_2]{\iff}(n_1n_2<0\,{\land}\,m_1n_2\,{\le}_{\mathbb{Z}}\,m_2n_1)\,{\lor}\,(n_1n_2>0\,{\land}\,m_2n_1\,{\le}_{\mathbb{Z}}\,m_1n_2)$.

How to prove that $\Big(\big(([a,b]\,{\le}\,[c,d])\,{\land}\,([c,d]\,{\le}\,[e,f])\big){\implies}([a,b]\,{\le}\,[e,f])\Big)$?

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    First, of course, you have to prove that this relation is well-defined... – 2017-01-05

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Actually if you restrict denominators to positive integers, you don't have to treat cases. If $\frac{a}{b}\leq\frac{c}{d}$ and $\frac{c}{d}\leq\frac{e}{f}$ then $ad\leq cb$ and $cf\leq ed$ so $adf\leq cbf\leq edb$ and $adf\leq edb$. Simplifying, we get $af\leq eb$, i.e. $\frac{a}{b}\leq\frac{e}{f}$.

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    Or if you prove that every $[a,b]$ contains an $(a',b')$ with $b'>0$, and you prove the relation is well-defined (why OP would have to do, anyway) then you could assume $b$ is positive. – 2017-01-05
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    @Thomas Andrews I supposed that since he is at the point where he wants to prove distributivity, he must already have proved that the relation is well-defined. – 2017-01-05
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    Right, my point was really that you need to do more than assert it is true for positive denominators, but rather that that you can assert a statement that each equivalent class contains such a pair with $b>0$. That's different from "restrict," which vaguely implies you are looking at a subset of the cases. You are not - you are asserting that every rational number has such a representative. – 2017-01-05
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    @Thomas Andrews Actually when I used the word "restrict", I had in mind defining $\mathbb{Q}$ as $\mathbb{Z}\times\mathbb{Z}_{>0}/\ominus$ instead of $\mathbb{Z}\times\left(\mathbb{Z}\backslash\{0\}\right)/\ominus$. In this sense, my answer was really meant as a hint for the OP. – 2017-01-05