$\newcommand{\A}{\mathscr{A}}\newcommand{\C}{\mathsf{C}}\newcommand{\T}{\mathcal{T}}\newcommand{\id}{\operatorname{id}}$Riehl, proposition $5.6.11$, from Category Theory in Context:
Suppose $\C$ is a cocomplete category and $\A$ is a category with all coequalisers. If $\A$ is monadic over $\C$, then $\A$ is cocomplete.
The proof is mostly left as an exercise. If you are unfamiliar with the notation used, I explain it in the second section. There is a very similar nLab article that is equally brief. I'll paraphrase her sketch:
By cocontinuity of equivalences, we can replace $\A$ by the category $\C^\T$ where $\T$ is the relevant monad, and $\C^\T$ must have all coequalisers.
Let $(A_i,\alpha_i)_{i\in I}\in\C^\T$ be any given family of algebras. The coproduct $(A,\alpha)$ of this family is defined to be the coequaliser at the right of the following diagram:
All objects in the fork are free $\T$-algebras. Using the adjunction $U^\T\vdash F^\T$ it is straightforward to show $(A,\alpha)$ is a coproduct. Then $\C^\T$ has all coproducts and coequalisers, thus all colimits.
Here, I assume $\kappa:\bigsqcup_i\T(A_i)\to \T(\bigsqcup_i A_i)$ is the unique arrow induced by the arrows $\T(j):\T(A_j)\to\T(\bigsqcup_i A_i)$, $j:A_j\hookrightarrow\bigsqcup_iA_i$ being the coproduct inclusion.
The trouble is, I can't make out what the inclusions arrows $(A_i,\alpha_i)\hookrightarrow (A,\alpha)$ are supposed to be! The nLab article says that it is not difficult to show $A=\bigsqcup_i A_i$, but what should $\alpha$ be? If we can construct this coequaliser by hand anyway, what's the point of assuming coequalisers exist? It just doesn't make sense to me.
My thoughts:
$(A,\alpha)$ is created as a coequaliser. We know nothing about it other than this, so to get an arrow $(A_j,\alpha_j)\to(A,\alpha)$ I have to route it through any of the objects in the coequaliser fork, almost certainly the free algebra $F^\T(\bigsqcup_i A_i)$. However, finding arrows into $F^\T$ is hard, since it is the left adjoint.
I need to find some arrow $\iota_j:A_j\to\T(\bigsqcup_i A_i)$ for every $j$ and: (i) hope $\iota_j$ lifts to an algebra homomorphism, (ii) hope that it is the right choice of inclusion arrow for the coproduct universal property. There is only one 'natural' choice of $\iota_j$ that I can see: $$\large\left(A_j\overset{j}{\hookrightarrow}\bigsqcup_i A_i\overset{\eta_{\sqcup_i A_i}}{\longrightarrow}\T\left(\bigsqcup_i A_i\right)\right)=\left( A_j\overset{\eta_{A_j}}{\longrightarrow}\T(A_j)\overset{\T(j)}{\longrightarrow}\T\left(\bigsqcup_i A_i\right)\right)$$
But unfortunately this isn’t necessarily a homomorphism. If this $\iota_j$ satisfies $\iota_j\alpha_j=\mu_{\sqcup_i A_i}\T(\iota_j)$, then that is iff: $$\begin{align}\T(j)\eta_{A_j}\alpha_j&=\mu_{\sqcup_i A_i}\T^2(j)\T(\eta_{A_j})\\&=\T(j)\mu_{A_j}\T(\eta_{A_j})\\&=\T(j)\end{align}$$Though $\alpha_j\eta_{A_j}=\id_j$, $\eta_{A_j}\alpha_j\neq\id_{\T(A_j)}$ in general. So this map isn't, as far as I can tell, guaranteed to be a homomorphism. There are also no other 'natural' choices, I think, for these inclusions - so I am stuck. By ‘natural’ I really mean to say: there just aren’t any other maps we can guarantee to exist, that I can think of.
I would really appreciate a concrete description of what the inclusion homomorphisms $(A_j,\alpha_j)\to(A,\alpha)$ are, and any hints / answers for showing $(A,\alpha)$ is indeed a coproduct would be most welcome.
Another point: as a left adjoint, $F^\T$ preserves all colimits, and $\C$ is cocomplete. Perhaps that is necessary here... but so far, Riehl has only suggested that we require $\C$ to have all coproducts - I'm not sure why $\C$ needs to have all colimits.
In case Riehl's notation isn't well-known, here is the definition of $\C^\T$:
For a monad $(\T,\eta,\mu)$ on a category $\C$, we define the Eilenberg-Moore category, a.k.a. category of $\T$-algebras, $\C^\T$ to have objects pairs $(c,\varsigma)$ where $c\in\C$ and $\varsigma:\T\C\to\C$ is some arrow in $\C$, the 'algebra structure map', which must satisfy the two conditions: $$\varsigma\eta_c=\id_c,\quad\varsigma\mu_c=\varsigma\T(\varsigma)$$Arrows in $\C^\T$ shall be "$\T$-algebra homomorphisms" $f:(c,\varsigma)\to(c',\varsigma')$ corresponding to arrows $f:c\to c'$ in $\C$ that satisfy the condition: $$f\varsigma=\varsigma'\T(f)$$The identity arrows and arrow composition are inherited from $\C$.
The adjunction she refers to is this: $F^\T:\C\to\C^\T$ is the functor assigning $c\mapsto(\T c;\mu_c)$ and $f:c\to c'$ is mapped to $\T(f)$ ($F^\T$ creates 'free algebras' and 'free maps'). $U^\T:\C^\T\to\C$ is the forgetful functor.
$\newcommand{\A}{\mathscr{A}}\newcommand{\C}{\mathsf{C}}\newcommand{\T}{\mathcal{T}}\newcommand{\id}{\operatorname{id}}$EDIT: About half a year later. There is indeed an easier proof available, that I worked out yesterday, after I was reminded of this post. Also, I realised my answer was slightly obscure (I got myself confused re-reading it yesterday, which is what prompted me to find the easier solution) so I would like to explain the notations a bit.
I'll just write the easier method now, really the rest is waffle but I think it was worth saying.
We should note that the unit $\eta$ for the adjunction $F^\T\dashv U^\T$ is the same as the unit $\eta$ for $\T$ (thankfully no conflict of notation!) and for any algebra $(c,\varsigma)$, the counit $\varepsilon_c$ will be $\mu_c:F^\T U^\T(c,\varsigma)=F^\T(c)\to(c,\varsigma)$.
Fix a family of algebras $(X_i,\xi_i)_{i\in I}$. I use $\overline{f}$ to denote the transpose of the arrow $f$ under the adjunction $F^\T\dashv U^\T$ and use standard identities to compute such transposes.
Families of algebra homomorphisms $(\varphi_j:(X_j,\xi_j)\to(c,\varsigma))_{j\in I}$
$\cong$
Families of arrows (in $\C$) $(\varphi_j:U^\T(X_j,\xi_j)\to U^\T(c,\varsigma))_{j\in I}$ satisfying: $$\varphi_j\circ U^\T(\varepsilon_{(X_j,\xi_j)})\equiv U^\T(\overline{\varphi_j})$$In $j\in I$.
$\cong$
Families of algebra homomorphisms $(\psi_j:F^\T U^\T(X_j,\xi_j)\to(c,\varsigma))_{j\in I}$ satisfying: $$\overline{\psi_j}\circ U^\T(\varepsilon_{(X_j,\xi_j)})\equiv U^\T(\psi_j)\\\iff\\\psi_j\circ F^\T U^\T(\varepsilon_{(X_j,\xi_j)})\equiv\psi_j\circ\varepsilon_{F^\T U^\T(X_j,\xi_j)}$$In $j\in I$
$\cong$
Algebra homomorphisms $\varphi:(X,\xi)\to(c,\varsigma)$ where $(X,\xi)$ is defined to be the coequaliser of the algebra homomorphisms $\sum_i\T(\xi_i)$ and $\sum_i\mu_{X_i}$ from: $$\sum_iF^\T U^\T F^\T U^\T(X_i,\xi_i)=\sum_i(\T^2X_i,\mu_{\T X_i})\to\sum_i(\T X_i,\mu_{X_i})=\sum_iF^\T U^\T(X_i,\xi_i)$$
Therefore, $(X,\xi)$ is a valid coproduct for the algebras $(X_i,\xi_i)_{i\in I}$. Stepping backwards through this proof with $\varphi:=\mathrm{Id}$ would find the inclusion homomorphisms $\varphi_j$ to be exactly $p\circ j'\circ\eta_{X_j}$, as equivalently stated by my older answer. This proof is nicer in that one hardly needs to think to find these inclusions, whereas half a year ago I was playing guessing games.
The below answer is absolutely correct, and a nice concrete calculation, but I would like to clarify it. I write things like $\sum_i(-)$ (of algebras), but one should be wary of using symbols that are not yet really defined. The whole point is that we don't have coproducts (yet!) available to use. So whenever you read something like $\sum_i(\T X_i,\mu_{X_i})$, we must read it as a more convenient notation for: $F^\T(\sum_i X_i)$. The two are literally equal, not just isomorphic, nice I have no other way (a priori) to construct coproducts.
On that note, stepping through the proof for why left adjoints preserve colimits, we can directly compute that: $$\sum_i\T\xi_i:\sum_i(\T^2X_i,\mu_{\T X_i})\to\sum_i(\T X_i,\mu_{X_i})$$Equals: $$\T\left(\sum_i\xi_i\right):F^\T\left(\sum_i\T X_i\right)\to F^\T\left(\sum_i X_i\right)$$And that: $$\sum_i\mu_{X_i}:\sum_i(\T^2X_i,\mu_{\T X_i})\to\sum_i(\T X_i,\mu_{X_i})$$Equals: $$\mu_{\sum_i X_i}\circ\T(\kappa):F^\T\left(\sum_i\T X_i\right)\to F^\T\left(\sum_i X_i\right)$$Where $\kappa$ is, as I originally suspected, the unique arrow $\sum_i\T X_i\to\T\left(\sum_i X_i\right)$ induced by the components $\T(j):\T X_j\to\T\left(\sum_i X_i\right)$.
I won't go into details, but you can verify these equalities by using the following (which I arrived at by stepping through the reason why left adjoints preserve colimits):
So the diagram used below to construct the coproduct is "exactly the same" as the one proposed by Riehl in the sense that, if $F^\T$ is used to construct coproducts of free algebras, then we get the above equalities.
Original self-answer:
A less opaque exposition is given here but it still omits some details that caused me a headache or two... A key point is my last observation that $F^\T$ preserves colimits. I will explain their exposition here for posterity and also for my own note-taking.
In their notation, we consider a family $(X_i,\xi_i)_{i\in I}$ of algebras. We know that $\sum_i X_i$ exists in $\C$, and that: $$\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)=F^\T\left(\sum_i X_i\right)\cong\sum_i(\T X_i,\mu_{X_i})$$Define the inclusions $j:X_j\to\sum_i X_j$, $j':(\T X_j,\mu_{X_j})\to\sum_i(\T X_i,\mu_{X_i})$, $j'':(\T^2 X_j,\mu_{\T X_j})\to\sum_i(\T^2 X_i,\mu_{\T X_i})$. We know that $T(j)$ corresponds to $j'$ under the isomorphism. This is because $F^\T$ is cocontinuous!
Now, we define an algebra $(X,\xi)$ by the coequaliser: $$\sum_i(\T^2 X_i,\mu_{\T X_i})\overset{\sum_i\mu_{X_i}}{\underset{\sum_i\T\xi_i}{\large\rightrightarrows}}\sum_i(\T X_i,\mu_{X_i})\cong\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)\overset{p}\longrightarrow(X,\xi)$$
We define the inclusions as follows:
So that settles my original question.
Now suppose $(\sigma,\varsigma)$ is some other $\T$-algebra with given homomorphisms $c_i:(X_i,\xi_i)\to(\sigma,\varsigma)$. Then $c_i\xi_i:(\T X_i,\mu_{X_i})\to(\sigma,\varsigma)$ are homomorphisms, inducing a unique homomorphism: $$g:\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)\to(\sigma,\varsigma)$$For which $g\circ\T(j)=c_j\circ\xi_j$ for all $j\in I$. Then: $$\sum_i(\T^2 X_i,\mu_{\T X_i})\overset{\sum_i\mu_{X_i}}{\underset{\sum_i\T\xi_i}{\large\rightrightarrows}}\sum_i(\T X_i,\mu_{X_i})\cong\left(\T\left(\sum_i X_i\right),\mu_{\sum_i X_i}\right)\overset{g}\longrightarrow(\sigma,\varsigma)$$Commutes; it is equivalent to check commutativity on every coordinate. Precomposing with an inclusion $j''$ gives $g\T(j)\mu_{X_j}$ versus $g\T(j)\T(\xi_j)$, that is, $c_j\circ\xi_j\circ\mu_{X_j}$ versus $c_j\circ\xi_j\circ\T(\xi_j)$. But: $\xi_j\mu_{X_j}=\xi_j\T(\xi_j)$ holds by definition of an algebra structure map $\xi_j$, so indeed the above fork commutes.
Thus $g=c\circ p$ for a unique homomorphism $c:(X,\xi)\to(\sigma,\varsigma)$. We can compute: $$\begin{align}c\circ h_j&=c\circ p\circ\eta_{\sum_i X_i}\circ j\\&=g\circ\T(j)\circ\eta_{X_j}\\&=c_j\circ\xi_j\circ\eta_{X_j}\\&=c_j\end{align}$$As desired. Moreover, if $\zeta:(X,\xi)\to(\sigma,\varsigma)$ is any other homomorphism with the property $\zeta\circ h_j=c_j$ for all $j\in I$, then: $$\zeta\circ p\circ\eta_{\sum_i X_i}\circ j\equiv g\circ\eta_{\sum_i X_i}\circ j,\,\forall j\in I$$Implies by the uniqueness of the coproduct arrow that: $$\zeta\circ p\circ\eta_{\sum_i X_i}=g\circ\eta_{\sum_i X_i}$$Then we use the fact that $\zeta,g$ and $p$ are all homomorphisms to get: $$\begin{align}g&=g\circ\mu_{\sum_i X_i}\circ\T(\eta_{\sum_i X_i})\\&=\varsigma\circ\T(g\circ\eta_{\sum_i X_i})\\&=\varsigma\circ\T(\zeta)\circ \T(p)\circ\T(\eta_{\sum_i X_i})\\&=\zeta\circ\xi\circ\T(p)\circ\T(\eta_{\sum_i X_i})\\&=\zeta\circ p\circ\mu_{\sum_i X_i}\circ\T(\eta_{\sum_i X_i})\\&=\zeta\circ p\end{align}$$But $g=\zeta\circ p$ implies $\zeta=c$ by uniqueness! Therefore, $c$ is the unique homomorphism with the property $c\circ h_j=c_j$ for all $j$.
That concludes $(X,\xi)$ is indeed a coproduct for $(X_i,\xi_i)_{i\in I}$.