= Solution
Use the elementary, or weak, <topos> axioms: <finite limits> and a <power object> $PA$ for every object $A$, naturally representing $\operatorname{Sub}(A\times X)$. No arbitrary products or coproducts are assumed. In particular $\Omega=P1$ is the <subobject classifier>. Inverse image of the membership relation defines $Pf:PB\to PA$ for $f:A\to B$.
Interchanging the factors in a relation gives natural bijections
$$
\mathcal E(A,PB)\cong\operatorname{Sub}(A\times B)\cong\mathcal E(B,PA).
$$
Thus the <contravariant power-object functor> $P:\mathcal E^{\mathrm{op}}\to\mathcal E$ has left adjoint $P^{\mathrm{op}}$. Its <adjunction unit> $\eta_A:A\to PPA$ sends a generalized element $a$ to the predicate on $PA$ given by membership of $a$.
This unit is monic. If $x,y:T\to A$ have the same image under $\eta_A$, membership of $x$ and $y$ agrees in every $T$-parameterized <subobject> of $A$. Apply this to the graph of $x$ in $A\times T$: membership of $x$ is true, so membership of $y$ is true, which means $y=x$. This argument uses parameterized relations, not an assumption that global elements detect morphisms.
The <functor> $P$ reflects isomorphisms. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality $PPf\,\eta_A=\eta_Bf$ together with monicity of $\eta_A$ shows that $f$ is monic. Regard $f$ as a <subobject> of $B$. Its inverse image along itself is the whole of $A$, just as for the whole <subobject> of $B$. Those two <subobjects> correspond to maps $1\to PB$; since $Pf$ is monic, they coincide. Therefore the <subobject> $f$ is all of $B$, and $f$ is an isomorphism.
We now prove the preservation condition for the <crude monadicity theorem>. A <reflexive pair> in $\mathcal E^{\mathrm{op}}$ is a <coreflexive pair> $f,g:B\rightrightarrows A$ in $\mathcal E$, with $rf=rg=1_B$ for some $r:A\to B$. Let $e:E\hookrightarrow B$ be its <equalizer>. For any <monomorphism> $m$, define $\exists_m$ on <power objects> by composing a parameterized <subobject> with $m$; no general image construction is needed because $m$ is monic. Inverse image followed by this extension obeys $Pm\,\exists_m=1$.
Here both $f,g$ are monic. For a parameterized <subobject> $S$ of $B$, an element $b$ belongs to $g^{-1}(f(S))$ exactly when $gb=fb'$ for some $b'\in S$. Applying $r$ gives $b=b'$, so this condition is precisely $b\in S$ and $fb=gb$. Consequently, as morphisms of <power objects>,
$$
Pf\,\exists_f=1_{PB},\qquad Pg\,\exists_f=\exists_e Pe,\qquad Pe\,\exists_e=1_{PE}.
$$
If $h:PB\to Z$ equalizes $Pf,Pg$, these identities give
$$
h=hPf\,\exists_f=hPg\,\exists_f=h\exists_e Pe.
$$
It therefore factors uniquely through $Pe$, which is a split <epimorphism> with section $\exists_e$. Since $fe=ge$, $Pe$ also equalizes $Pf,Pg$. Hence $Pe$ is their <coequalizer>. This proves explicitly that <power objects turn coreflexive equalizers into coequalizers>.
Finite <equalizers> in $\mathcal E$ give all <coequalizers> of reflexive pairs in $\mathcal E^{\mathrm{op}}$, and we have shown that $P$ preserves them and reflects isomorphisms. The <crude monadicity theorem> now yields
$$
\boxed{P\text{ is monadic},\qquad\mathcal E^{\mathrm{op}}\simeq\mathcal E^{PP}.}
$$
Fix a set $I$ of cardinality $\kappa$. If $\mathcal E$ has $I$-indexed <products in a category>, its <Eilenberg-Moore category> for the double-power <monad> has them, since the forgetful <functor> creates existing limits. The displayed equivalence then gives $I$-products in $\mathcal E^{\mathrm{op}}$, hence $I$-<coproducts in a category> in $\mathcal E$.
For the converse, suppose those coproducts exist. The diagonal $\Delta:\mathcal E\to\mathcal E^I$ is a <logical functor>, because all the elementary topos structure in $\mathcal E^I$ is pointwise, and it has left adjoint $\coprod_I$. Apply <power-object monadicity> to both toposes. Under their monadic identifications, $\Delta^{\mathrm{op}}$ is the lift of $\Delta$ to the double-power <Eilenberg-Moore categories>. The base <functor> $\Delta$ is a right adjoint to $\coprod_I$, so the <adjoint lifting theorem for monad algebra functors> gives a left adjoint to its algebra lift. Its required reflexive <coequalizers> exist in $\mathcal E^{\mathrm{op}}$, since they are opposites of finite <equalizers> in $\mathcal E$. Taking opposites gives a right adjoint to $\Delta$, which is exactly the $I$-product <functor>. Thus the <products and coproducts of fixed cardinality in a topos> satisfy
$$
\boxed{\mathcal E\text{ has }\kappa\text{-products}\iff\mathcal E\text{ has }\kappa\text{-coproducts}.}
$$
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