Reference request for category theory works which quickly prove the theorem which generalises the 1st isomorphism theorem for groups/rings/...

If you want a categorical proof that encompasses nonabelian groups and rings-without-identity, then you need to work with concepts such as "normal subobject" that are (I think) not really part of the category theory that most mathematicians routinely encounter. You may also find that a proof is not particularly enlightening at this level of generality. I think this might explain why a simultaneous generalization does not appear early in textbooks.

I'll consider the formulation of the first isomorphism theorem which states that for any homomorphism $f: A to B$, $ker(f): K to A$ is a normal subobject, and the image of $f$ in $B$ is isomorphic to the quotient of $A$ by $K$. The formalism is roughly as follows: We work in a category in which every morphism has a kernel and a cokernel. Then any morphism $f: A to B$ factors into a composition of $coker(ker(f)): A to A/K$, $m: A/K to N$, and $ker(coker(f)): N to B$, and the diagram is defined up to unique isomorphism by the universal properties of kernel and cokernel. For the first part of the theorem, one typically defines normal subobject to be a kernel (although there are alternative definitions involving congruences), so we reduce to the question of the image. Here we have a conflict of terminology, since the "image" is often defined in categories as the kernel of the cokernel, and the first isomorphism theorem interprets image set-theoretically. Given this alternative interpretation, the theorem amounts to the assertion that $m$ and $ker(coker(f))$ are monomorphisms. Here we encounter another problem, since as far as I know, $m$ does not need to be a monomorphism in general.

At this point, we need to check that the categories we like have kernels and cokernels. The kernel of a group homomorphism $f:A to B$ is (the inclusion of) the preimage of the identity, and the cokernel of $f$ is given by taking the normal hull $N$ of $f(A)$ in $B$, and taking the quotient group $B/N$. We get our theorem because the maps of $f(A)$ into $N$ and $N$ into $B$ are inclusions, hence monomorphisms. A similar argument works for rings-without-identity. For abelian groups and vector spaces, the map $m: f(A) to N$ is an isomorphism. For fields, there are no kernels, so the first isomorphism theorem doesn't make sense.