What does the Yoneda lemma say for the identity functor and finite sets?

The Yoneda Lemma in the stated form tells you about natural transformations between the functors $\text{Hom}(-,A)$ and $F$ from $\mathbf{Fin}$ to $\mathbf{Set}$. The crucial point is that $\text{Hom}(-,A)$ is contravariant, and this requires the functor $F$ to be contravariant, too. But the $F$ you chose, namely $\text{id}_\mathbf{Fin}$ is covariant.

Let's just replace it by a real contravariant functor, the preimage functor $\cal P^{-1}:\mathbf{Fin}→\mathbf{Set}$, which assigns to a set $S$ its power set and to a function $f:S\to T$ the set map $\cal PT→\cal PT$ which gives the preimage of every subset. Now the Yoneda Lemma claims that $$\text{Nat}(\text{Hom}(-,\{a,b,c\}),\cal P^{-1}-)\cong \cal P\{a,b,c\}$$ So choose an element $B$ of $\cal P\{a,b,c\}$. Now if $f:S\to\{a,b,c\}$ is a set map, we can assign to $f$ the preimage $f^{-1}[B]$ which is an element of $\cal P^{-1}S$

Of course if you want to keep the $\text{id}_\mathbf{Fin}$ functor, you are invited to replace the contravariant hom-functor by $\text{Hom}(A,-)$ and see what happens.

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Edit

Your solution is right. The covariant Yoneda Lemma establishes a bijection $$\text{Nat}(\text{Hom}(A,-),F-)\cong FA$$ where the element $e\in FA$ corresponds to the natural transformation $\sigma_e$ which sends a morphism $f:A→B$ to $(Ff)(e)$.
In the case $F=\text{id}_\mathbf{Set}$ and $A=\{a,b,c\}$, the element $a\in A$ gives the transformation $\alpha:\ \mathbf{Set}(A,-)\ \longrightarrow\ (-)$ which maps $f:A→B$ to $f(a)$.
Yoneda tells us that this is natural, that means for $g:B→C$ $$α(\mathbf{Set}(A,g)(f))=g(α(f))$$ which is nothing else than $$(g\circ f)(a)=g(f(a))$$ which is what you said.
So in this case, Yoneda doesn't give us much new insight, but there are cases where it is really useful.