Here is a nice trick to construct an example. But maybe there are more naturally occuring examples. I feel like there should be a better way to explain the construction, but I don't know how for now.
The core of the idea is the following observation:
The category of suplattices (poset with arbitrary suprema) is monadic over Set (the power set being the monad) so it can be represented by a product sketch. It has all colimits.
The category of sets X endowed with a "successor" function $f:X \to X$ is also sketchable, in fact it is a presheaf category, it also has all colimits.
But, the category of suplattices $S$ endowed with a successor function (not a morphism) $f:S \to S$ does not have all colimits. These are called ZF-algebras in algebraic set theory and the initial ZF-algebra can be identified with the class of all sets (the supremum being identified with the union of set and the function $f$ with $X \mapsto \{X\}$). See for example page 4-5 of An outline of algebraic set theory by Awodey for more details).
This category of ZF-algebras is (probably) not going to be sketchable directly (at least, not with a 'realized' sketch) but we can exploit this fact nonetheless.
The trick is to consider the category of triples $(X,A,B)$, where $X$ is a set, $A$ and $B$ are "subsingleton" (i.e. sets that are either empty or singletons) and such that if $A = \{*\}$ then $X$ has a suplattice structure, and if $B = \{*\}$ then $X$ has a function $f:X \to X$. Think of $A$ and $B$ as "toggles" that activate additional structure on $X$. Functions $(X,A,B) \to (X',A',B')$ only exist if $A \leqslant A'$ and $B \leqslant B'$ and are functions $X \to X'$ which are suplattice morphisms if $A=1$ and preserve the "successor" function if $B=1$.
This is sketchable, informally because you can use a sketch that does not contain any object where both $A=\{*\}$ and $B=\{*\}$ at the same time (no axiom involve both structure at the same time). But using colimit of representable, we will get objects where $A = B =\{*\}$, hence ZF-algebras and arrive in a world where colimits can be large.
Let me try to give an explicit description of the sketch. I'll describe the opposite category of the sketch, so that I can work with full subcategory of the category of models which I find more convenient. It will have for objects:
- For each set $X$ an object $T_X$ corresponding to $(X,\emptyset,\emptyset)$.
- For each set $X$ an object $F_X$ corresponding to $(\mathbb{N} \times X,\emptyset,1)$ with the obvious succesor function.
- An object $S_X$ for each set $X$ corresponding to $(\mathcal{P}(X),1,\emptyset)$ with its suplattice structure.
They have all their morphisms between the corresponding models, so basically the $T_X$ have all functions between sets as morphisms, the $S_X$ have their suplattice morphisms between them, the $F_X$ have the morphisms of $\mathbb{N}$-sets between them and the only morphisms between these classes are the maps from the $T_X$ to the $F_Y$ and $S_Z$ that corresponds respectively to functions from $X$ to $\mathbb{N} \times Y$ and from $X$ to $\mathcal{P}(Z)$.
If we only put the obvious (co)limit condition on each of these classes separately, then the category of models would a category of triples of a suplattice $S$, an $\mathbb{N}$-set $F$ and a set $T$ together with functions from $S$ to $T$ and from $F$ to $T$. That is not enough yet to break co-completeness.
But we have one more card to play:
First one removes the condition that $S_\emptyset$ and $T_\emptyset$ are initial, which expands the category above a bit by replacing "suplattice" and "$\mathbb{N}$-set" by the categories obtained by freely adding initial objects to the category of suplattices and $\mathbb{N}$-sets (see the edit below for more details).
Then we add new marked cocone, implementing the condition that $S_\emptyset \coprod T_1 = S_{1}$ and $F_\emptyset \coprod T_1 = F_{1}$ through the obvious maps. It is easy to check that these are indeed colimits in the category of representables (i.e. in the sketch), so this is still a realized sketch.
Now, the triples as above are model of this new sketch if as soon as the suplattice $S$ is not the freely added initial object, then the map $S \to T$ is a bijection and similarly, as soon as the $F$ is not the freely added initial object, then the map $F \to T$ is a bijection.
This gives us exactly the category of triples $(X,A,B)$ described above ($A$ and $B$ just remembering if the "suplatice" or the "$\mathbb{N}$-Set" are the freely added initial object or not).
So, in the resulting category of models, if you try to construct the coproduct of $(\{0,1\},1,\emptyset)$ with $(\emptyset,\emptyset,1)$ you are constructing $(V,1,1)$ where $V$ is the initial ZF-algebra, which is always large (it is in bijection with the class of all sets as mentioned before).
Edit: let me details a side construction that I'm using which I think will clarify a bit what I'm doing here.
Assume that I have a "realized" sketch $T$, which I see as a full subcategory of its category of models. For simplicity, I'm assuming that T is stable under finite coproducts, and that all cocone in T that are colimits in the category of models are marked cocone of the sketch.
I now consider the sketch T' with the same underlying category as T, but where I have removed from the list of marked cocone all the empty ones (i.e. the cocone that would specify an initial object). I claim that the category of T' model identify with the category of T-model together with a new, freely added initial object given by the empty model.
Indeed, as all special cocone of T' are non-empty, one easily see that the empty model (sending each ovject of $T$ to the empty set) is a model of T'.
Now given any non-empty model X of T', i.e. such that X(a) is inhabited for some object $a$. In T', if I write $e$ for the initial object of $T'$, we have $a \coprod e \simeq a$ and it corresponds to a marked cocone $a,e \to a$. This mean that the unique map $e \to a$, induce a map $X(a) \to X(e)$ such that the map $X(a) \to X(a) \times X(e)$ is a bijection, in particular, the projection map $X(a) \times X(e) \to X(a)$ being a right inverse of it is also a bijection. But as $X(a)$ is inhabited, this implies that $X(e)=\{*\}$, and hence that $X$ is not justs a $T'$ model but actually a $T$ model as it is also compatible to the empty cocone of $T$.
Given that all $T$-model are non-empty (as $X(e)=\{*\}$) we have that a $T'$ model is either the empty model or a $T$-model. Hence the result.
