What do identities mean in $\mathrm{Set}^\mathrm{op}$? Since $\mathrm{Set}$ has finite coproducts, thus we may consider models of equational theories in the opposite category $\mathrm{Set}^\mathrm{op}$. The result is basically that function symbols $f : X^n \rightarrow X$ in the signature are interpreted as functions $X \rightarrow nX$ in $\mathrm{Set}.$
Which is pretty cool, but what do identities mean in this context?
For a couple of simple examples: how would we interpret the identities expressing commutativity, or associativity?
 A: Commutativity: $f:X\times X \to X$ is commutative if $f\circ s=f$ where $s:X\times X\to X\times X$ is the "swap map" that exchanges the roles of the two canonical projection maps.
Analogously, $g:X\to X+X$ would be "co-commutative" if $\sigma\circ g=g$ where $\sigma:X+X\to X+X$ is a "swap" that exchanges the roles of the two canonical injections.
With elements, if we represent $A+B$ as $(A\times\{0\})\cup (B\times\{1\})$ what this means that whenever $g(x)=(y,0)$ we must also have $g(x)=(y,1)$ and vice versa. This is impossible unless $X=\varnothing$, so in $\mathbf{Set}^{\rm op}$ you can only have a commutative identity in this trivial case.

Associativity: With similar element-tracing arguments, it seems that a "co-associative" $g:X\to X+X$ would require a kind of "mock idempotence": whenever $g(x) = (y,i)$ (for $i\in\{0,1\}$) it must hold that $g(y)=f(y,i)$ too.
This is possible in non-degenerate cases, such as for example $g:\mathbb Z\to\mathbb Z+\mathbb Z$ given by
$$ g(x) = \begin{cases} (42,0) & x\text{ is even} \\ (|x|,1) & x\text{ is odd} \end{cases} $$
but still it is not nearly as interesting as in $\mathbf{Set}$, because a "co-associative" map just consists of a partition of $X$ into to sets $A$ and $B$ together with idempotent maps $A\to A$ and $B\to B$. So the two sides of the sum don't really interact usefully, like they can in an ordinary associative operation.

In general: In $\mathbf{Set}$, an expression describes a "machine" where we must put an element into each of the named input slots; then we can crank the handle and a single result comes out at the root of the expression. An identity between two expressions asserts that the two machines behave the same (viewed from the outside).
Dually, an expression in $\mathbf{Set}^{\rm op}$ describes a "machine" where we supply a single input value at the root; after we crank the handle an output will drop out at exactly one of the named output slots.
In $\mathbf{Set}^{\rm op}$, a constant letter (or any other expression with no free variables) must represent a machine where, after we crank the handle, a value drops out at exactly one of no named output slots. This is impossible unless the handle is stuck. Formally, a constant letter would need to be interpreted by a function $X\to\varnothing$, and such a thing exists only if $X$ is empty.
Therefore, if you want to interpret $x+(-x)=0$ in $\mathbf{Set}^{\rm op}$ you run into the problem that "$0$" cannot be interpreted at all unless $X=\varnothing$, in which case everything is boring.
