I offer the following summary/interpretation of Broodryk's results.

Let $\mathcal{A}$ be a category of algebras, by which I mean a category equipped with a (strictly) monadic functor $U : \mathcal{A} \to \textbf{Set}$.
Let $F : \textbf{Set} \to \mathcal{A}$ be left adjoint to $U$ and let $T = U F$.
Broodryk implicitly only considers finitary algebraic theories (i.e. $\mathcal{A}$ is locally finitely presentable and $U : \mathcal{A} \to \textbf{Set}$ preserves filtered colimits) but I think his main results also apply to infinitary algebraic theories.

First, as a warm up:

**Lemma.**
If a category has an initial object that is also a strict terminal object (i.e. every morphism with domain a terminal object is an isomorphism), then every object in that category is initial/terminal/zero. ◼

A trivial observation, to be sure, but turning it around tells us something about the set $T (0)$ of constants in the algebraic theory:

**Proposition.**
If $\mathcal{A}$ has a strict terminal object, then either:

* $T (0)$ is empty, or
* $T (0)$ has at least two elements, or
* $\mathcal{A}$ is trivial.

*Proof.*
$T (0)$ is the underlying set of the initial object $F (0)$ in $\mathcal{A}$.
If $T (0)$ has exactly one element, then $F (0)$ is both an initial object and a strict terminal object in $\mathcal{A}$, in which case $\mathcal{A}$ is trivial. ◼

Next:

**Proposition.** If $\mathcal{A}$ has codisjoint binary products then $T (0)$ is not empty.

*Proof.* 
Let $A$ and $B$ be objects in $\mathcal{A}$.
The product $A \times B$ is codisjoint if the following is a pushout square in $\mathcal{A}$:
$$\require{AMScd}
\begin{CD}
A \times B @>{\pi_2}>> B \\
@V{\pi_1}VV @VVV \\
A @>>> 1
\end{CD}$$
Consider the case $A = B = F (0)$.
Assume $T (0)$ is empty.
Then $U (F (0) \times F (0)) \cong T (0) \times T (0)$ is also empty, hence both projections $F (0) \times F (0) \to F (0)$ are isomorphisms.
But the pushout of an isomorphism is an isomorphism, so that implies $F (0) \to 1$ is an isomorphism, which is a contradiction.
So $T (0)$ is not empty. ◼

**Corollary.**
If $\mathcal{A}$ is coextensive, then either:
* $T (0)$ has at least two elements, or
* $\mathcal{A}$ is trivial.

*Proof.*
A coextensive category has both a strict terminal object and codisjoint binary products. ◼

So far so good – after all, a rig has two distinguished elements.
But it is not obvious how to extract binary operations from the hypothesis that $\mathcal{A}$ is coextensive.
What Broodryk found is that, if $\mathcal{A}$ is coextensive, then in the coproduct $A + F (0) \times F (0)$ in $\mathcal{A}$ there is a binary operation that, when applied to pairs of elements in the image of the coproduct insertion $\iota_1 : A \to A + F (0) \times F (0)$, yield an element from which we can recover the original elements by applying the morphism $[\textrm{id}_A, \pi_j] : A + F (0) \times F (0) \to A$, thereby exhibiting $A + F (0) \times F (0)$ as the binary product $A \times A$.
Essentially, it is a pairing operation.

**Theorem.**
$\mathcal{A}$ is coextensive if and only if the following conditions hold:

* There exist a regular cardinal $\kappa$ ($\le \lambda$ if $F (0) \times F (0)$ admits a generating set of $\le \lambda$ elements), an element $t \in T (2 + \kappa)$, and $\vec{e} \in T (0)^\kappa$ and $\vec{e}{}' \in T (0)^\kappa$, such that the following equations hold in every $A$ in $\mathcal{A}$,
$$\begin{aligned}
t (x, x', \vec{e}) & = x \\
t (x, x', \vec{e}{}') & = x'
\end{aligned}$$
where $x$ and $x'$ are arbitrary elements of $A$ and (by abuse of notation) we have identified $t$ with the operation $U (A)^{2 + \kappa} \to U (A)$ it corresponds to, and $\vec{e}$ and $\vec{e}{}'$ with their images in $U (A)^\kappa$ under the unique morphism $F (0) \to A$.

* For every $A$ in $\mathcal{A}$, there is an isomorphism $\delta_A : A \times A \to A + F (0) \times F (0)$ in $\mathcal{A}$ (whose underlying map $U (\delta_A)$ is) given by
$$\delta_A (x, x') = t (\iota_1 (x), \iota_1 (x'), \iota_2 (\vec{e}, \vec{e}{}'))$$
where $t, \vec{e}, \vec{e}{}'$ are as above and (by abuse of notation) we have identified $(\vec{e}, \vec{e}{}')$ with the corresponding element of $(T (0) \times T (0))^\kappa$ under the natural bijection $T (0)^\kappa \times T (0)^\kappa \to (T (0) \times T (0))^\kappa$.

* Moreover, $\delta_A (x, x) = \iota_1 (x)$, where $\iota_1 : A \to A + F (0) \times F (0)$ is the coproduct insertion.　◻

I will sketch just the construction of $t$.
Consider the following diagram in $\mathcal{A}$:
$$\begin{CD}
F (0) @<{\pi_1}<< F (0) \times F (0) @>{\pi_2}>> F(0) \\
@V{!_A}VV @V{\iota_2}VV @VV{!_A}V \\
A @<<{[\textrm{id}_A, \pi_1]}< A + F (0) \times F (0) @>>{[\textrm{id}_A, \pi_2]}> A \\
@| @V{\eta_A}VV @| \\
A @<<{\pi_1}< A \times A @>>{\pi_2}> A 
\end{CD}$$
The two squares in the top half are pushout squares (always).
The morphism $\eta_A : A + F (0) \times F (0) \to A \times A$ is defined by the universal property of $A \times A$, and it is an isomorphism if $\mathcal{A}$ is coextensive.
Consider the case where $A = F (2)$.
Let $x$ and $x'$ be the two distinguished generators of $F (2)$.
Then $(x, x')$ is an element of $U (F (2) \times F (2))$, and $U (\eta_X) : U (F (2) + F (0) \times F (0)) \to U (F (2) \times F (2))$ is a bijection, so it has a preimage $\bar{t}$.

The hypothesis on $\kappa$ ensures there is an effective epimorphism $F (2 + \kappa) \to F (2) + F (0) \times F (0)$ and we can arrange for the first two distinguished generators of $F (2 + \kappa)$ to be mapped to the image under $\iota_1 : F (2) \to F (2) + F (0) \times F (0)$ of the two distinguished generators of $F (2)$ and the remaining $\kappa$ generators to be mapped into the image of $\iota_2 : F (0) \times F (0) \to F (2) + F (0) \times F (0)$.
Let $(\vec{e}, \vec{e}{}')$ be a preimage in $T (0)^\kappa \times T (0)^\kappa$ of the images of the $\kappa$ generators, considered as an element of $U (F (2) + F (0) \times F (0))^\kappa$.

Then, $[\textrm{id}_A, \pi_1] (\bar{t}) = t (x, x', \vec{e})$ by homomorphicity, and $[\textrm{id}_A, \pi_1] (\bar{t}) = \pi_1 (\eta_X (\bar{t})) = x$ because the bottom left square commutes, so $t (x, x', \vec{e}) = x$ as required.
Similarly, $t (x, x', \vec{e}{}') = x'$.
Since we have proved the claim for the two distinguished generators of $F (2)$, the claim follows for all pairs of elements in all algebras.

The morphism $\delta_A : A \times A \to A + F (0) \times F (0)$ is simply the inverse of $\eta_A : A + F (0) \times F (0) \to A \times A$.

*Remark.*
We have $\vec{e} = \vec{e}{}'$ if and only if $\mathcal{A}$ is trivial.
So this is another way of seeing that $T (0)$ has at least two elements if $\mathcal{A}$ is coextensive and not trivial.

**Example.**
For the theory of commutative rigs, we can take $\kappa = 2$, $t (x, x', y, y') = x y + x' y'$, $\vec{e} = (1, 0)$, and $\vec{e}{}' = (0, 1)$.
The point is that, in $A$,
$$\begin{aligned}
t (x, x', 1, 0) & = x \\
t (x, x', 0, 1) & = x'
\end{aligned}$$
and, in $A \otimes_\mathbb{N} (\mathbb{N} \times \mathbb{N})$ (the change in operator precedence by switching from $+$ to $\otimes$ here is confusing, but I digress),
$$\delta_A (x, x') = t (x \otimes 1, x' \otimes 1, 1 \otimes (1, 0), 1 \otimes (0, 1)) = x \otimes (1, 0) + x' \otimes (0, 1)$$
i.e. $\delta_A : A \times A \to A \otimes_\mathbb{N} (\mathbb{N} \times \mathbb{N})$ is the obvious natural isomorphism. ◼

It seems difficult to come up with an example of an algebraic theory satisfying all three of the conditions in the theorem.
The first condition – the existence of $t, \vec{e}, \vec{e}{}'$ satisfying the given equations – is easy enough to arrange.
(We can just freely add such $t, \vec{e}, \vec{e}{}'$ to any existing algebraic theory and get a new one!)
The hard part seems to be in the second and third conditions.
Broodryk remarks that the first condition alone is enough to guarantee many of the properties of coextensivity.