= Solution
A <monad> on $\mathcal C$ is an endofunctor $T$ with <natural transformations> $\eta:1\to T$ and $\mu:T^2\to T$ satisfying
$$
\mu\,T\mu=\mu\,\mu T,\qquad \mu\,T\eta=1_T=\mu\,\eta T.
$$
An <algebra for a monad> is $(A,\alpha:TA\to A)$ with $\alpha\eta_A=1_A$ and $\alpha T\alpha=\alpha\mu_A$. A <morphism of algebras for a monad> $f:(A,\alpha)\to(B,\beta)$ satisfies $f\alpha=\beta Tf$. These form the <Eilenberg-Moore category> $\mathcal C^T$. Its <free algebra functor> is $FX=(TX,\mu_X)$, and its <adjunction> to the forgetful <functor> $U$ has the explicit bijection
$$
\mathcal C^T(FX,(C,\gamma))\cong\mathcal C(X,C),\qquad
h\longmapsto h\eta_X,\qquad u\longmapsto\gamma Tu.
$$
Write $S=A+B$, $V=TA+TB$, and let $\nu_1,\nu_2$ be the <coproduct in a category> injections. Set
$$
r=F(\alpha+\beta),\qquad s=\mu_SF\kappa: FV\rightrightarrows FS,
\qquad \kappa=[T\nu_1,T\nu_2].
$$
Here $\mu_S:FTS\to FS$ is an algebra <morphism>, with underlying multiplication $T^2S\to TS$. For an algebra $(C,\gamma)$, an algebra <morphism> $h:FS\to(C,\gamma)$ corresponds to $u:S\to C$, and the transposes of $hr,hs$ are respectively
$$
u(\alpha+\beta),\qquad h\kappa=\gamma Tu\,\kappa.
$$
For the second formula, $(\mu_ST\kappa)\eta_V=\kappa$ by <naturality> of $\eta$ and the <monad> unit law. Thus $hr=hs$ exactly when, writing $u=[a,b]$,
$$
a\alpha=\gamma Ta,\qquad b\beta=\gamma Tb.
$$
These say precisely that $a$ and $b$ are <morphisms of algebras for a monad>. Consequently any <coequalizer> $q:FS\to Q$ of $r,s$ represents pairs of algebra <morphisms> out of $(A,\alpha)$ and $(B,\beta)$, proving the <coproduct presentation for monad algebras>:
$$
\boxed{Q\cong(A,\alpha)+(B,\beta)\quad\text{in }\mathcal C^T.}
$$
Its two algebra injections have underlying <morphisms> $q\eta_S\nu_1$ and $q\eta_S\nu_2$; the equivalence just proved verifies both the algebra equations and their universal property.
The displayed pair is a <reflexive pair>, with common section
$$
t=F(\eta_A+\eta_B):FS\to FV.
$$
Indeed $rt=1_{FS}$ by the algebra unit laws, while
$$
\kappa(\eta_A+\eta_B)=\eta_S,\qquad st=\mu_SF\eta_S=1_{FS}.
$$
Now suppose $\mathcal C$ has all finite <colimits> and $T$ preserves <reflexive coequalizers>. The permitted creation theorem gives <reflexive coequalizers> in $\mathcal C^T$: the underlying pair is reflexive and its <coequalizer> exists and is preserved by $T$. The preceding construction therefore gives binary algebra <coproducts in a category>. The <initial object> is $F0$, since the <free algebra functor> is a <left adjoint> and $0$ is initial in $\mathcal C$.
For any algebra pair $a,b:X\rightrightarrows Y$, the augmented pair
$$
[a,1_Y],[b,1_Y]:X+Y\rightrightarrows Y
$$
is a <reflexive pair>, with common section the injection of $Y$, and has exactly the same coequalizing <morphisms> as $a,b$. Hence all <coequalizers> exist in $\mathcal C^T$. The <construction of finite colimits from coproducts and reflexive coequalizers> now gives
$$
\boxed{\mathcal C^T\text{ has all finite colimits}.}
$$
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