On the proof that $1_{\textbf{Set}}$ is representable Example 4.1.4 says:

Consider $H^1:\textbf{Set}\to\textbf{Set}$, there $1$ is the one-element set. Since a map from $1$ to a set $B$ amounts to an element of $B$, we have $$H^1(B)\cong B$$ for each set $B$. It is easily verified that this isomorphism is natural in $B$, so $H^1$ is isomorphic to the identity functor $1_{\textbf{Set}}$.

In my understanding, the isomorphism mentioned at the end is constructed as follows. Let $\eta$ be the natural transformation from $H^1$ to $1_{\textbf{Set}}$ defined by $\eta_A(\star\mapsto a)= a$. That is, $\eta_A:H^1(A)\to 1_{\textbf{Set}}(A)$ is the arrow (in the category of sets) that sends the map $\star\mapsto a$ to $a$ (where $Ob(1)=\{\star\})$. If $f:A\to B$ is an arrow in the category of sets, one sees that the diagram
$$
\require{AMScd}
\begin{CD}
H^1(A) @>{H^1(f)}>> H^1(B);\\
@VVV @VVV \\
1_{\textbf{Set}}(A)=A @>{1_{\textbf{Set}}(f)=f}>> 1_{\textbf{Set}}(B)=B;
\end{CD}$$
commutes. Now one can similarly define the natural transformation $\epsilon$ from $1_{\textbf{Set}}$ to $H^1$ by $\epsilon_A(a)=(\star\mapsto a)$ and prove that both vertical compositions of $\epsilon$ and $\eta$ are identity transformations.
Now for my questions:
0) Is my proof correct?
1) I haven't used that $H^1(B)\cong B$ at all. Why does the text refer to this isomorphism? Is there another way to prove the desired claim using this fact?
2) I don't remember the notion of an isomorphism being "natural in" $B$ defined before, and there are no results (when using ctrl+F in the cited document) for "natural in" that appear before this example. What's the precise definition of this notion? I think Definition 1.3.12 is used here. See On the definition of $F(A)$ being isomorphic to $G(A)$ naturally in $A$
 A: Your proof is indeed correct. As for the second question: note that a natural transformation $\alpha : F \Rightarrow G$ is an isomorphism if and only if every component is an isomorphism in the target category of $F$ and $G$. Thus, instead of describing an explicit inverse for $\eta$, you could have just observed that the arrow
$$
\eta_B : f \in \mathsf{Set}(1,B) \mapsto f(\ast) \in B
$$ 
is an isomorphism for any set $B$. This is what the author is referring to, and what you are indirectly proving.
Edit: for completeness sake and to clear some doubts, I'll include the following result,

Proposition. Let $F,G : C \to D$ be two functors and $\alpha : F \Rightarrow G$  a natural transformation. The following statements are equivalent:
  
  
*
  
*$\alpha$ is a natural isomorphism.
  
*$\alpha_c : Fc \to Gc$ is an isomorphism for all $c \in \mathsf{ob}(C)$.
  

Proof. Suppose that $\alpha$ is an isomorphism, so that there exists $\beta : G \Rightarrow F$ with $\alpha \beta = 1_G$ and $\beta \alpha = 1_F$. Then, for each object $c \in \mathsf{ob}(C)$ we have 
$$
\alpha_c\beta_c = (\alpha\beta)_c = (1_F)_c = 1_{Gc}
$$
and similarly $\beta_c\alpha_c = 1_{Fc}$. Hence $\alpha_c$ is an isomorphism with inverse $\beta_c$.
Reciprocally, suppose that $\alpha_c$ is an isomorphism for each $c \in \mathsf{ob}(C)$, and note $\beta_c$ the inverse of $\alpha_c$. 
If $f : x \to y$ is an arrow in $C$, by naturality of $\alpha_c$ we obtain
$$
\alpha_{y}Ff = Gf \alpha_x.
$$
Right-composing by $\beta_y$ yields $Ff = \beta_y Gf\alpha_x$ and left-composing by $\beta_x$ we conclude that
$$
Ff\beta_x = \beta_yGf.
$$
This is to say that the collection of arrows $(\beta_c : Gc \to Fc)_{c \in \mathsf{ob}(C)}$ assembles into a natural transformation. Since by hypothesis each $\beta_c$ is the inverse of $\alpha_c$, it follows that $\beta \alpha = 1_F$ and $\alpha \beta  = 1_G$. Therefore $\alpha$ is a natural isomorphism.$\ \square$
