Non-trivial dual fields 
A dual field is a structure made out a set $S$, two binary operations $*$ and $\circ$ and two distinct special elements $1^*$ and $1^\circ$.


$S \cup 1^*$ form an Abelian group under $*$ with $1^*$ as the identity. ($1^\circ$ is an absorbing element under $*$)
$S \cup 1^\circ$ form an Abelian group under $\circ$ with $1^\circ$ as the identity. ($1^*$ is an absorbing element under $\circ$).
In addition, they both follow $(a \cdot b)\circ c = (a \circ c) \cdot (b \circ c)$ and $(a \circ b)\cdot c = (a \cdot c) \circ (b \cdot c)$.

Do any non-trivial dual fields exist?
If not, can it be proven?
Note: the trivial dual field has two elements $\{0,1\}$ with the operations $\max$ and $\min$.
 A: Towards a contradiction, let's take $x,y \in S$.
From distributivity, we have
$$(1^\circ * x) \circ y = (1^\circ \circ y) * (x \circ y)$$
Since $1^\circ$ is $*$-absorbative, we see
$$y = y * (x \circ y)$$
Since $*$ has group structure (unless possibly $1^\circ$ is involved), we conclude one of the following:

*

*If $1^\circ$ is nowhere to be seen, then we can cancel $y$ from each side and get $1^* = x \circ y$. But since we assumed $x,y \in S$, this means $S \cup \{1^\circ\}$ is not closed under $\circ$, contradicting group-ness.


*If instead one of the above terms is $1^\circ$, it must be $x \circ y$. This is because we assumed $x,y \in S$, so neither can be $1^\circ$. But then we would have
$y = y * 1^\circ$, which, by absorbativity again, means $y = 1^\circ$. A contradiction.
Thus, we must have $S = \emptyset$, leaving the trivial case as the only one.
(As a brief remark - if you don't insist on $1^* \neq 1^\circ$, $\{1\}$ with only trivial operations is another example. But that's even less exciting than what you called the trivial example.)

I hope this helps ^_^
