Actually, it is a partial order. Given natural numbers $a,b$, we can conclude from $a\vert b\wedge b\vert a$ that $a=b.$ Your counterexample doesn't work, because, while everything divides $0$, $0$ only divides itself.
A maximum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\leq a\wedge d\vert b$. This would require that $a$ be the greatest element of $\Bbb Z^-$ and that $b$ be a natural number that all natural numbers divide. Can you think of what that would be? Note that if there is a maximum element, then it is the only maximal element.
A minimum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\geq a\wedge b\vert d$. Then $a$ would have to be the least element of $\Bbb Z^-$, which is not possible.
Given any $(a,b)$, we have that $(a-1,b)\odot(a,b)$, so there is always a different element preceding it. Thus, there are no minimal elements, either.
To determine whether or not $\odot$ totally orders $T$, we must determine whether for distinct $(a,b),(c,d)\in T$ we need have $(a,b)\odot(c,d)$ or $(c,d)\odot(a,b)$. It shouldn't be difficult to see that this need not hold (I leave it to you to find an example), so that $\odot$ does not totally order $T$, after all.
Edit: Well done in showing that it isn't a total order, and in finding the maximum element. It is in fact maximum (not just maximal) since for each $(c,d)\in T$ we have $c\leq-1$ and $d\vert 0$. Partial orders may or may not have a maximum or a minimum element, and may or may not have maximal or minimal elements. For an example of a partial order with a maximum element, take any set $A$, any $b\notin A$ and define a partial order $\precsim$ on $A\cup\{b\}$ by $c\precsim c$ for all $c$, and $a\precsim b$ for all $a\in A$. Then $b$ is the maximum element, and each $a\in A$ is incomparable to the others, so minimal, but not minimum if $A$ has multiple elements.
To elaborate on the antisymmetry of your example, let's suppose that $m,n\in\Bbb N$ such that $m\vert n\wedge n\vert m$. Hence, $$m=nv\wedge n=mw$$ for some $v,w\in\Bbb N$. If $m=0$, then so is $n$. If $m\neq 0$, then since $m=nv=mvw$, then $vw=1$, which is only possible for $v=w=1$ since $v,w\in\Bbb N$. Thus, again, $m=n$. Does that clear things up?
Best Answer
If $S$ is a partially ordered set in which is $s$ is a minimal element, then $\forall x\in S(x\leq s\implies x=s)$.
Since $R$ is a total order and $s$ is minimal, it follows that $s$ is in fact a minimum.
Now to answer your question.
Hint: Suppose $s'\in S$ is also a minimum. Relate it do $s$ and try to conclude that $s=s'$.
What you wrote is good enough for an answer, but you're asking for a 'more symbolic' answer, so here it is.
Full answer:
As stated above, since $R$ is a partial order with a minimum element $s\in S$, then $\forall x\in S(x\leq s\implies x=s)$.
But $R$ is a total order, thus $\forall x\in S(x\leq s\lor s\leq x)$, therefore, since $s$ is minimal $\forall x\in S(x=s\lor s\leq x)$, i.e., $\forall x\in S(s\leq x)$, that is, $s$ is a minimum of $S$ with respect to the order $R$.
Now suppose $s'$ is also a minimum of $S$ with respect to $R$, then, by definition of $s'$ being a minimum, it holds that $s'\leq s$. On the other hand, $s$ is a minimum too, so $s\leq s'$. From $s'\leq s\land s\leq s'$ it follows that $s=s'$. This proves that all minimum elements are the same, that is, there is a unique minimum element.
It's probably good to note that the fact that $R$ is a total order is only necessary to prove that a minimum exists.
On any poset, if there is a minimum, the reasoning above proves it is unique.