I'm not sure how to define "classically" here (Hensel's $p$-adics hail from around 1890s), but the Hilbert symbol gives a way to do this without $p$-adics. (See Chapter 12 of my book, http://quatalg.org.) In a nutshell, multiplying by an even power of $p$, we may assume without loss of generality that $a,b$ are integers not divisible by $p^2$. We may suppose also that $b$ is not divisible by $p$: if necessary, either swap $a,b$ or replace $b$ by $-ab/p^2$ (taking the standard generators $i,ij$ in place of $i,j$). Then $(a,b\,|\,\mathbb{Q})$ ramifies at an odd prime $p$ if and only if:

- $p \nmid ab$, or
- $p \mid a$ and ($b$ is not a square modulo $p$).

For $p=2$, there is something more elaborate modulo $8$ (see
(12.4.16)), and of course there is a simple criterion at the infinite place (over $\mathbb{R}$).

If you like, you an also interpret this as whether or not the conic $ax^2+by^2=1$ has a solution modulo an arbitrary power of $p$ (hiding the $p$-adics, but not very well yet). With the above normalizations, using Hensel's lemma you can see that this is equivalent to the above.

This isn't a very conceptual criterion! But it is useful.

To your second question: yes, if $\mathcal{O}$ has basis $1,i,j,k$ (not necessarily 'standard' generators), then it has a *reduced discriminant* (square root of the discriminant) $\mathrm{discrd}(\mathcal{O})=\pm \mathrm{trd}((ij-ji)k)$ and this is exactly the product of the primes that ramify (chapter 15). For example, in the algebra $B := (-1,3\,|\,\mathbb{Q})$, the order $\mathbb{Z} \oplus \mathbb{Z} i \oplus \mathbb{Z} j \oplus \mathbb{Z} k$ where $k=(1+i+j+ij)/2$ is maximal, and
$ \mathrm{trd}((ij-ji)k)=6 $
showing that $B$ is ramified at $2,3$.

Of course, there's much more to say here!