Using the given Identity Axioms:
(E1) $\forall x \exists y (x=y)$
If (E1) is not true, then Axiom I1:
$\forall x (x = x)$
will not be true. By Contradiction, (E1) is true.
(E2) It looks like we can not show this with Only the Identity Axioms; There may be more Axioms listed earlier in the Text Book, which, together with Axioms I1 & I4, must be used in (E2) here.
I think that (E2) is trying to state that variable $x$ can be substituted by variable $y$ in a statement, Provided that $y$ does not occur in the statement.
Eg $x=zx+1$ is equivalent to $y=zy+1$, but $x=yx+1$ is not equivalent to $y=yy+1$
UPDATE:
The first Part of (E2) can be shown via "Contradiction", with DeMorgans Laws:
$\varphi(x) \Leftrightarrow \exists x (x=y \land \varphi(y))$
When LHS is true, RHS must be true. Assume RHS is not true; then the negation is true.
$\lnot (\exists x (x=y \land \varphi(y)))$
$\forall x ( \lnot (x=y) \lor \lnot(\varphi(y)))$
$( \forall x \lnot (x=y) \lor \forall x \lnot(\varphi(y)))$
By (E1), $\forall x \lnot (x=y)$ is not true.
Thus, $\forall x \lnot(\varphi(y))$ must be true.
By given LHS, there is $\varphi(x)$, then $\forall x \lnot(\varphi(y))$ is not true.
It is a "Contradiction" here.
Thus Negation of RHS is not true.
Hence, RHS is true.
Here $\varphi(x)$ must not have $y$, because that might make this argument invalid:
$\varphi(x) : x=z+1$ can have at least one Integer Solution, and $\varphi(x)$ does not have $y$.
$\varphi(y) : y=z+1$ is same.
But:
$\varphi(x) : x=y+1$ can have at least one Integer Solution, but $\varphi(x)$ does have $y$.
$\varphi(y) : y=y+1$ is not the same and has no Integer Solutions.
Alternate Proof of (E2) with I4:
We could take I4 with $n=1$ to get:
$\forall x,y (x=y \rightarrow (\varphi(x) \rightarrow \varphi(y))) $
$\forall x,y ((x=y \land \varphi(x)) \rightarrow (\varphi(y))) $
$\forall x,y ((\varphi(x) \land x=y) \rightarrow (\varphi(y))) $
$\forall x (\varphi(x) \rightarrow \exists y (x=y \land \varphi(y))) $
Best Answer
Note that induction is equivalent to well-ordering (more generally to well-foundedness). Namely, removing the induction axiom, a model of $\sf PA$ is well-ordered if and only if it satisfies the (second-order) induction axiom.
But well-ordering is equivalent to "there is no infinite decreasing chain". Finally, since in $\sf PA$ every non-zero element has a predecessor, this means that well-ordering is equivalent to stating that no element has infinitely many elements smaller than itself.
And this should be fairly straightforward to state using $\exists^\infty$.