Before addressing your questions I will write here my favorite version of the first isomorphism theorem. As others have commented, you need notions of quotients, images and kernels before even attempting to enunciate such a result. There's more than one way to do this (for example, additive categories). Here we are going to work with concrete categories. Recall the notion of images.
Definition [kernel]: Let $f : X \to Y$ be a function. The kernel of $f$ is the set $\{(a,b) \in X \times X \mid f(a)=f(b)\}$.
The notion of kernel as defined here is simply the kernel pair of $f$, that is, the pullback of $X \xrightarrow{f} Y \xleftarrow{f} X$.
Definition [concrete quotients and congruences]: Let $(C,U)$ be a concrete category and $X$ an object of $C$. A concrete quotient of $X$ is an epimorphism $\pi : X \to Y$ such that $U(\pi)$ is epi and for every object $Z$ of $C$ and every function $f : U(Y) \to U(Z)$, the following are equivalents:
- There exists a morphism $f' : Y \to Z$ such that $U(f') = f$.
- There exists a morphism $g : X \to Z$ such that $U(g) = f \circ U(\pi)$.
The set $\ker(U(\pi))$ is called a congruence on $X$.
If you prefer, you can define concrete quotients as equivalence classes instead. Note that this notion of quotient coincides with topological quotients, for instance, while the usual notion of quotients (that is, epimorphisms) does not. In essence, concrete quotients allow you to complete diagrams in the base category by looking at the underlying diagrams in $Set$. A congruence on an object $X$ is essentially an equivalence relation on $U(X)$ with an associated concrete quotient of $X$. Observe, however, that congruences need not to arise only from $U(\pi)$ for $\pi$ a concrete quotient.
Theorem [the First Isomorphism Theorem]: Let $(C,U)$ be a concrete category, where $C$ is complete and $U$ is continuous. Let $q : X \to Z$ be a morphism in $C$ such that $\ker(U(q))$ is a congruence on $X$. Then the morphism $m : X/\ker(U(q)) \to Z$ (such that $q = m \circ \pi_q$) is the image of $q$.
Proof : First of all we must verify that $m$ is a monomorphism. Let $x,y \in U(X)$ and $[x],[y]$ their equivalence classes regarding $\ker(U(q))$. If $U(m)([x])=U(m)([y])$, then $(U(\pi_q) \circ U(m))(x)=(U(\pi_q) \circ U(m))(y)$, hence $U(\pi_q \circ m)(x)=U(\pi_q \circ m)(y)$, which implies $U(q)(x)=U(q)(y)$. Therefore $(x,y) \in \ker(U(q))$ and $[x]=[y]$. $U(m)$ is mono, hence $m$ is as well ($U$ is faithful).
Now let $m' : Y \to Z$ be a monomorphism and $h : X \to Y$ be a morphism such that $q= m' \circ h$. we wish to prove the existence of $f : X/\ker((U(q))) \to Y$ such that $m = m' \circ f$. If $(x,y) \in \ker(U(q))$, then $U(q)(x)=U(q)(y)$, hence $U(h)(x)=U(h)(y)$ (since $U$ is continuous and $m'$ is mono, $U(m')$ is mono). By the definition of concrete quotients, there exists a morphism $f : X/\ker((U(q))) \to Y$ such that $h = f \circ \pi_q$. Since $m' \circ h = q = m \circ \pi_q$, we have $m' \circ f \circ \pi_q = m \circ \pi_q$. Since $\pi_q$ is epi, we have $m' \circ f = m$.
Note that, in particular, this isomorphism theorem is valid on the category of topological spaces (with the obvious forgetful functor to $Set$)! What is the problem here? I'll leave that as an exercise.
Now, addressing the questions:
So is it correct that I shouldn't be trying to prove the isomorphism theorems in category theory?
That is not correct. However, you should be aware of the fact that general categories might not always have the structure/properties you need to talk about certain concepts. In that case, however, you should still be able to consider a particular class of categories in which you can prove your desired results.
Is it correct that instead, the isomorphism theorems are kind of like "interfaces", which justify talking about algebraic structures (among other structuers) in terms of structure preserving morphisms?
I'm not sure if I understand this question. The fact that homomorphisms between algebraic structures satisfy the isomorphism theorems is certainly a good reason to talk about structure preserving functions (instead of non-structure preserving functions) in certain scenarios. However, in other structures, where the theorem might not be valid (topological spaces for example), it is still "better" to consider structure preserving functions than simply general functions.
