Algebroids are particularly interesting structures: they are basically categories enriched over $K$-Vect for some field $K$. This means the hom-sets are all vector spaces, and composition is "bilinear" in a certain sense. (Let's just focus on when $K$ is a field for now rather than an arbitrary ring.)
Most examples I can think of are equivalent to some subset of matrices with the usual addition and multiplication rules, as long as we are willing to be creative and allow "infinite matrices" to exist. In general, for any $n$ and $m$, the set of $n \times m$ matrices forms an algebroid. The union of any such sets also forms an algebroid, as long as matrix compositions exist when expected (meaning if we have $5 \times 4$ and $4 \times 3$ matrices, we must also have $5 \times 3$ matrices). This is also true if $n$ and $m$ are arbitrary infinite cardinals, with the caveat that only finitely many elements of each column of the matrix can be nonzero.
So, we can ask if this is basically "what algebroids are," in the following sense:
Let's say that the category $\text{Mat}^+_K$ is basically an extension of the usual $\text{Mat}_K$, but with the objects as all possible cardinals rather than only natural numbers. For objects $\kappa, \lambda$, the morphisms from $\kappa \to \lambda$ are (possibly infinite) matrices of size $\kappa \times \lambda$ (treating these cardinals as initial ordinals), with finitely many nonzero coefficients in each column, taking values in $K$.
Naive Pre-Question: is every possible $K$-algebroid equivalent to a subcategory of $\text{Mat}_K^+$?
Now, the answer to this first question is "no," because we can have, for instance, a disjoint union of $\text{Mat}_K^+$ with itself, which is a $K$-algebroid but not equivalent to $\text{Mat}_K^+$. This would be like having, for instance, "red" and "blue" matrices, where the product of red and blue matrices is undefined. So, to salvage the spirit of this question, we can note that $\text{Mat}_K^+$ is a skeleton category of $K$-Vect, which has infinitely many copies of $\text{Mat}_K^+$ with all possible linear transformations (and thus isomorphisms, when they exist). So, we can get to the better question:
Real Question: is every possible $K$-algebroid equivalent to a subcategory of $\text{Vect}_K$?
If the category is small, then we can indeed define a functor $\Phi : \mathcal{C} \to \operatorname{Vect}_K$ which sends an object $X$ of $\mathcal{C}$ to the vector space $\bigoplus_{Y \in \operatorname{ob}(\mathcal{C})} \operatorname{Hom}_{\mathcal{C}}(Y, X)$. The image of a morphism $f \in \operatorname{Hom}(X, X')$ will be the unique $K$-linear map such that $g \in \operatorname{Hom}(Y, X)$ will be sent to $f \circ g \in \operatorname{Hom}(Y, X')$. It is straightforward to check that this defines a functor; moreover, it is easy to see that it is faithful since $f$ is uniquely determined by where $\Phi(f)$ sends $\operatorname{id}_X \in \operatorname{Hom}(X, X)$.