A toy example, using the Yoneda lemma:

Claim: There are two canonical bialgebra structures (the “additive” and “multiplicative” structures) on $k[x]$, and one of them (the additive one) in fact makes it a Hopf algebra.

Proof 1: (Calculation.) Write down the formulas; check the axioms! This isn't an especially long calculation, but it's a bit tedious; while seeing the formulas is nice, checking the axioms isn't (to my taste) especially enlightening.

Proof 2: (Abstract.) “Bialgebra” = “comonoid in ($k$-Alg,$\otimes$)”. We know $k[x]$ is the free $k$-algebra on one generator, so there's a natural isomorphism $\mathrm{Hom}(k[x],A) \cong A$, for any $k$-algebra $A$. So $\mathrm{Hom}(k[x],A)$ is naturally an algebra — so it has two natural monoid structures, + and $\cdot$, and under + it's moreover a group. By Yoneda, these must correspond to two comonoid structures on $k[x]$, and the one corresponding to + must be Hopf!

Now, what I really like about this proof is that it still connects closely to the computations. By the way that the Yoneda lemma works, you can read off what the two coalgebra structures actually are; but now you don't have to check the axioms, since you already know they hold! Also, you now know there'll be a “co-distributive law” connecting the two, which you might never have thought of just from the first approach… And also, this gives a way of looking for bialgebra structures on other algebras: look at what they classify/represent!

This shows up, I think, a lot of the power of abstract approaches. They put formulas and calculations into a bigger picture; they can help you do interesting calculations, while letting you skip tedious ones; and they can suggest calculations you might not have thought of doing otherwise. But (as you can probably guess from that) I love calculation too: I wouldn't want either without the other. If abstract nonsense is the garden, concrete computations are the flowers.