Is there a "purely algebraic" proof of the finiteness of the class number?
Yes, there exist purely algebraic conditions on a Dedekind domain which hold for all rings of integers in global fields and which imply that the class group is finite.
For a finite quotient domain $A$ (i.e., all non-trivial quotients are finite rings), a non-zero ideal $I\subseteq A$ and a non-zero $x\in A$, let $N_{A}(I)=|A/I|$ and $N_{A}(x)=|A/xA|$. Also define $N_{A}(0)=0$.
Call a principal ideal domain $A$ a basic PID if the following conditions are satisfied:
$A$ is a finite quotient domain,
for each $m\in\mathbb{N}$,$$\#\{x\in A\mid N_{A}(x)\leq m\}>m$$ (i.e., $A$ has “enough elements of small norm”),
there exists a constant $C\in\mathbb{N}$ such that for all $x,y\in A$, $$N_{A}(x+y)\leq C\cdot(N_{A}(x)+N_{A}(y))$$ (i.e., $N_{A}$ satisfies the “quasi-triangle inequality”).
Theorem. Let $A$ be a basic PID and let $B$ be a Dedekind domain which is finitely generated and free as an $A$-module. Then $B$ has finite ideal class group.
For the proof, see here.
It is easy to verify that $\mathbb{Z}$ and $\mathbb{F}_q[t]$ are basic PIDs, so the ring of integers in any global field satisfies the hypotheses of the above theorem (using the non-trivial fact that rings of integers in global fields are finitely generated over one of these PIDs).
More generally, one can take the class of overrings of Dedekind domains which are finitely generated and free over some basic PIDs. Since it is known that an overring of a Dedekind domain with finite class group also has finite class group, this gives a wider class of algebraically defined Dedekind domains (including $S$-integers like $\mathbb{Z}[\frac{1}{p}]$) with finite class group.
Added: The second condition for basic PIDs can be relaxed to: there exists a constant $c\in\mathbb{N}$ such that for each $m\in\mathbb{N}$, $$ \#\{x\in A\mid N_{A}(x)\leq c\cdot m\}\geq m. $$
A nice account of Dedekind's `greatest common divisor' of two algebraic number is given in Hecke's book, 'Lectures on the Theory of Algebraic Numbers'. You wont of course see an algebraic proof there.
The closest I know of an answer to your question is given by Stickelberger's theorem on ideal class annihilators. In Ireland and Rosen at the end of Chapter 14 you can even find an algebraic proof that the class group of $\mathbb{Q}(\sqrt{-p}), p\equiv 3 \mod 4$ is annihilated by the classical $(N-R)/p$. In fact it is the class number, but I don't know that you can prove this algebraically.
Of course this only applies to abelian extensions of $\mathbb{Q}$, so it doesn't really come close to answering the general question!