I cannot really come up with a specific interesting examples of categories like this which would not come from universal algebra

Everything comes from universal algebra!

Joking aside, I mean that we have to be more precise about this requirement before the question can be answered. I will describe a class of examples and let others tell me whether this class comes from universal algebra.

Let $L$ be a bounded lattice considered as a category. Us usual, the statement that $L$ is being considered as a category means that the objects of the category are the elements of $L$ and whenever $a\leq b$ in $L$ we demand that $\textrm{Hom}(a,b)$ have a unique element, otherwise $\textrm{Hom}(a,b)$ is empty.

There are no nontrivial pairs of parallel arrows in categories like this. Thus, the only (co)equalizers are the identity maps. While every morphism is both a monomorphism and an epimorphism, the only regular monomorphisms or regular epimorphisms are the identity maps. The only regular subobject of $a\in L$ is $\textrm{id}_a\colon a\to a$. The opposite category is a category of the same type.

That is, in a sufficiently (co?)regular category $\mathbf{C}$, an object $X$ is subdirectly irreducible if, in the opposite category, for any family $X_i\rightarrowtail X$ of regular subobjects with the induced map $\bigsqcup X_i\to X$ (regular?) epi, one of the $X_i$ must be $X$ itself.

If you test the example against this definition you get that each regular subobject $X_i\rightarrowtail X$ is an instance of $\textrm{id}_X\colon X\to X$. The induced map $\bigsqcup X_i\to X$ will also be an identity map, hence it will be regular epi. We will have one of the $X_i$ equal to $X$ itself, since all $X_i$ must be $X$ itself. The conclusion is that, under the given definition, all elements of $L$ are subdirectly irreducible.

If you drop the demand that $X_i\rightarrowtail X$ be a regular monomorphism and only require that it be a monomorphism, an element of $L$ will be subdirectly irreducible if and only if it is meet-irreducible in $L$. In that case, if you start with an algebraic lattice $L$, where every element is a complete meet of meet-irreducibles, you will get that every object of $L$ is a subdirect product of subdirectly irreducible objects of $L$.

I cannot really come up with a specific interesting examples of categories like this which would not come from universal algebra

One has to be more precise about this requirement before the question can be answered. I would say that a small category is a $2$-sorted partial algebra $\langle \textrm{Ob}, \textrm{Mor}; \circ, \textrm{Id}, \textrm{dom}, \textrm{cod}\rangle$, where $\circ\colon \textrm{Mor}\times \textrm{Mor}\to \textrm{Mor}$ is a partial operation while $\textrm{Id}\colon \textrm{Ob}\to \textrm{Mor}$, $\textrm{dom}\colon \textrm{Mor}\to \textrm{Ob}$, and $\textrm{cod}\colon \textrm{Mor}\to \textrm{Ob}$ are total operations. The study of small categories then may be viewed as a part of universal algebra (if your universal algebra allows multisorted partial algebras). I will describe a class of examples and let others tell me whether this class comes from universal algebra.

Let $L$ be a bounded lattice considered as a category. Us usual, the statement that $L$ is being considered as a category means that the objects of the category are the elements of $L$ and whenever $a\leq b$ in $L$ we demand that $\textrm{Hom}(a,b)$ have a unique element, otherwise $\textrm{Hom}(a,b)$ is empty.

There are no nontrivial pairs of parallel arrows in categories like this. Thus, the only (co)equalizers are the identity maps. While every morphism is both a monomorphism and an epimorphism, the only regular monomorphisms or regular epimorphisms are the identity maps. The only regular subobject of $a\in L$ is $\textrm{id}_a\colon a\to a$. The opposite category is a category of the same type.

That is, in a sufficiently (co?)regular category $\mathbf{C}$, an object $X$ is subdirectly irreducible if, in the opposite category, for any family $X_i\rightarrowtail X$ of regular subobjects with the induced map $\bigsqcup X_i\to X$ (regular?) epi, one of the $X_i$ must be $X$ itself.

If you test the example against this definition you get that each regular subobject $X_i\rightarrowtail X$ is an instance of $\textrm{id}_X\colon X\to X$. The induced map $\bigsqcup X_i\to X$ will also be an identity map, hence it will be regular epi. We will have one of the $X_i$ equal to $X$ itself, since all $X_i$ must be $X$ itself. The conclusion is that, under the given definition, all elements of $L$ are subdirectly irreducible.

If you drop the demand that $X_i\rightarrowtail X$ be a regular monomorphism and only require that it be a monomorphism, an element of $L$ will be subdirectly irreducible if and only if it is meet-irreducible in $L$. In that case, if you start with an algebraic lattice $L$, where every element is a complete meet of meet-irreducibles, you will get that every object of $L$ is a subdirect product of subdirectly irreducible objects of $L$.

I cannot really come up with a specific interesting examples of categories like this which would not come from universal algebra

Everything comes from universal algebra!

Joking aside, I mean that we have to be more precise about this requirement before the question can be answered. I will describe a class of examples and let others tell me whether this class comes from universal algebra.

Let $L$ be a bounded lattice considered as a category. Us usual, the statement that $L$ is being considered as a category means that the objects of the category are the elements of $L$ and whenever $a\leq b$ in $L$ we demand that $\textrm{Hom}(a,b)$ have a unique element, otherwise $\textrm{Hom}(a,b)$ is empty.

There are no nontrivial pairs of parallel arrows in categories like this. Thus, the only (co)equalizers are the identity maps. While every morphism is both a monomorphism and an epimorphism, the only regular monomorphisms or regular epimorphisms are the identity maps. The only regular subobject of $a\in L$ is $\textrm{id}_a\colon a\to a$. The dual category is a category of the same type.

That is, in a sufficiently (co?)regular category $\mathbf{C}$, an object $X$ is subdirectly irreducible if, in the opposite category, for any family $X_i\rightarrowtail X$ of regular subobjects with the induced map $\bigsqcup X_i\to X$ (regular?) epi, one of the $X_i$ must be $X$ itself.

If you test the example against this definition you get that each regular subobject $X_i\rightarrowtail X$ is an instance of $\textrm{id}_X\colon X\to X$. The induced map $\bigsqcup X_i\to X$ will also be an identity map, hence it will be regular epi. We will have one of the $X_i$ equal to $X$ itself, since all $X_i$ must be $X$ itself. The conclusion is that, under the given definition, all elements of $L$ are subdirectly irreducible.

I cannot really come up with a specific interesting examples of categories like this which would not come from universal algebra

Everything comes from universal algebra!

Joking aside, I mean that we have to be more precise about this requirement before the question can be answered. I will describe a class of examples and let others tell me whether this class comes from universal algebra.

Let $L$ be a bounded lattice considered as a category. Us usual, the statement that $L$ is being considered as a category means that the objects of the category are the elements of $L$ and whenever $a\leq b$ in $L$ we demand that $\textrm{Hom}(a,b)$ have a unique element, otherwise $\textrm{Hom}(a,b)$ is empty.

There are no nontrivial pairs of parallel arrows in categories like this. Thus, the only (co)equalizers are the identity maps. While every morphism is both a monomorphism and an epimorphism, the only regular monomorphisms or regular epimorphisms are the identity maps. The only regular subobject of $a\in L$ is $\textrm{id}_a\colon a\to a$. The opposite category is a category of the same type.

That is, in a sufficiently (co?)regular category $\mathbf{C}$, an object $X$ is subdirectly irreducible if, in the opposite category, for any family $X_i\rightarrowtail X$ of regular subobjects with the induced map $\bigsqcup X_i\to X$ (regular?) epi, one of the $X_i$ must be $X$ itself.

If you test the example against this definition you get that each regular subobject $X_i\rightarrowtail X$ is an instance of $\textrm{id}_X\colon X\to X$. The induced map $\bigsqcup X_i\to X$ will also be an identity map, hence it will be regular epi. We will have one of the $X_i$ equal to $X$ itself, since all $X_i$ must be $X$ itself. The conclusion is that, under the given definition, all elements of $L$ are subdirectly irreducible.

If you drop the demand that $X_i\rightarrowtail X$ be a regular monomorphism and only require that it be a monomorphism, an element of $L$ will be subdirectly irreducible if and only if it is meet-irreducible in $L$. In that case, if you start with an algebraic lattice $L$, where every element is a complete meet of meet-irreducibles, you will get that every object of $L$ is a subdirect product of