[proofplan]
We compare the rank functions of the two matroids in each asserted isomorphism on the common ground set $E\setminus\{e\}$. The proof uses the rank formula for the dual matroid together with the rank formulas for deletion and contraction, including the loop convention for contraction. Once the two rank functions agree on every subset of the common ground set, the identity map on $E\setminus\{e\}$ is an isomorphism of matroids.
[/proofplan]
[step:Fix the common ground set and rank notation]
Let
\begin{align*}
E_0 := E\setminus\{e\}
\end{align*}
denote the common ground set after deleting or contracting $e$. For any matroid $N$, let
\begin{align*}
r_N:2^{E(N)} \to \mathbb{N}\cup\{0\}
\end{align*}
denote its rank function, where $2^{E(N)}$ is the power set of its ground set.
We use the dual rank formula: for every matroid $N$ with ground set $F$ and every subset $X\subset F$,
\begin{align*}
r_{N^*}(X)=|X|-r_N(F)+r_N(F\setminus X).
\end{align*}
We also use the deletion and contraction rank formulas. If $x\in F$ and $A\subset F\setminus\{x\}$, then
\begin{align*}
r_{N\setminus x}(A)=r_N(A)
\end{align*}
and
\begin{align*}
r_{N/x}(A)=r_N(A\cup\{x\})-r_N(\{x\}).
\end{align*}
The contraction formula is valid whether $x$ is a loop or a nonloop, since $r_N(\{x\})$ is either $0$ or $1$.
[/step]
[step:Compute the rank function of the dual of the deletion]
Let $A\subset E_0$ be arbitrary. The matroid $M\setminus e$ has ground set $E_0$, so the dual rank formula applied to $N=M\setminus e$ and $F=E_0$ gives
\begin{align*}
r_{(M\setminus e)^*}(A)=|A|-r_{M\setminus e}(E_0)+r_{M\setminus e}(E_0\setminus A).
\end{align*}
By the deletion rank formula,
\begin{align*}
r_{M\setminus e}(E_0)=r_M(E_0)=r_M(E\setminus\{e\}).
\end{align*}
Also, since $E_0\setminus A=E\setminus(A\cup\{e\})$, the deletion rank formula gives
\begin{align*}
r_{M\setminus e}(E_0\setminus A)=r_M(E\setminus(A\cup\{e\})).
\end{align*}
Therefore
\begin{align*}
r_{(M\setminus e)^*}(A)=|A|-r_M(E\setminus\{e\})+r_M(E\setminus(A\cup\{e\})).
\end{align*}
[/step]
[step:Compute the rank function of the contraction of the dual]
Let $A\subset E_0$ be arbitrary. The contraction $M^*/e$ has ground set $E_0$, and the contraction rank formula applied in $M^*$ gives
\begin{align*}
r_{M^*/e}(A)=r_{M^*}(A\cup\{e\})-r_{M^*}(\{e\}).
\end{align*}
Apply the dual rank formula in $M$ to the subset $A\cup\{e\}\subset E$:
\begin{align*}
r_{M^*}(A\cup\{e\})=|A\cup\{e\}|-r_M(E)+r_M(E\setminus(A\cup\{e\})).
\end{align*}
Since $A\subset E\setminus\{e\}$, we have $|A\cup\{e\}|=|A|+1$, hence
\begin{align*}
r_{M^*}(A\cup\{e\})=|A|+1-r_M(E)+r_M(E\setminus(A\cup\{e\})).
\end{align*}
Applying the same dual rank formula to $\{e\}$ gives
\begin{align*}
r_{M^*}(\{e\})=1-r_M(E)+r_M(E\setminus\{e\}).
\end{align*}
Substitution into the contraction formula yields
\begin{align*}
r_{M^*/e}(A)=\bigl(|A|+1-r_M(E)+r_M(E\setminus(A\cup\{e\}))\bigr)-\bigl(1-r_M(E)+r_M(E\setminus\{e\})\bigr).
\end{align*}
Cancelling the terms $1$ and $-r_M(E)$ gives
\begin{align*}
r_{M^*/e}(A)=|A|-r_M(E\setminus\{e\})+r_M(E\setminus(A\cup\{e\})).
\end{align*}
This is exactly the expression obtained for $r_{(M\setminus e)^*}(A)$.
[guided]
The goal is to compare $M^*/e$ with $(M\setminus e)^*$ on the same ground set $E_0=E\setminus\{e\}$. Since matroids are determined by their rank functions, it is enough to compute $r_{M^*/e}(A)$ for an arbitrary subset $A\subset E_0$ and compare it with the formula already obtained for $r_{(M\setminus e)^*}(A)$.
The contraction rank formula in the matroid $M^*$ says that contraction by $e$ has rank
\begin{align*}
r_{M^*/e}(A)=r_{M^*}(A\cup\{e\})-r_{M^*}(\{e\}).
\end{align*}
This formula is the right tool because contraction is the operation appearing on the right-hand side, and it reduces the problem to evaluating ranks in the dual matroid $M^*$.
We now expand each dual rank using the dual rank formula in the original matroid $M$. For the subset $A\cup\{e\}\subset E$, the complement in $E$ is $E\setminus(A\cup\{e\})$, so
\begin{align*}
r_{M^*}(A\cup\{e\})=|A\cup\{e\}|-r_M(E)+r_M(E\setminus(A\cup\{e\})).
\end{align*}
Because $A\subset E\setminus\{e\}$, the element $e$ is not already in $A$, so $|A\cup\{e\}|=|A|+1$. Therefore
\begin{align*}
r_{M^*}(A\cup\{e\})=|A|+1-r_M(E)+r_M(E\setminus(A\cup\{e\})).
\end{align*}
For the singleton $\{e\}$, the complement in $E$ is $E\setminus\{e\}$, and the dual rank formula gives
\begin{align*}
r_{M^*}(\{e\})=1-r_M(E)+r_M(E\setminus\{e\}).
\end{align*}
Substituting both expansions into the contraction formula gives
\begin{align*}
r_{M^*/e}(A)=\bigl(|A|+1-r_M(E)+r_M(E\setminus(A\cup\{e\}))\bigr)-\bigl(1-r_M(E)+r_M(E\setminus\{e\})\bigr).
\end{align*}
The constants $1$ and $-r_M(E)$ cancel with their opposites after distributing the minus sign. Hence
\begin{align*}
r_{M^*/e}(A)=|A|-r_M(E\setminus\{e\})+r_M(E\setminus(A\cup\{e\})).
\end{align*}
This is precisely the same formula obtained for $r_{(M\setminus e)^*}(A)$. Since $A\subset E_0$ was arbitrary, the two rank functions agree on every subset of the common ground set $E_0$.
[/guided]
[/step]
[step:Deduce the first isomorphism from equality of ranks]
For every subset $A\subset E_0$, the previous two computations show
\begin{align*}
r_{(M\setminus e)^*}(A)=r_{M^*/e}(A).
\end{align*}
Both matroids have ground set $E_0$. Therefore the identity map
\begin{align*}
\operatorname{id}_{E_0}:E_0\to E_0
\end{align*}
preserves rank on every subset, so it is a matroid isomorphism. Hence
\begin{align*}
(M\setminus e)^*\cong M^*/e.
\end{align*}
[/step]
[step:Compute the rank function of the dual of the contraction]
Let $A\subset E_0$ be arbitrary. The matroid $M/e$ has ground set $E_0$, so applying the dual rank formula to $N=M/e$ and $F=E_0$ gives
\begin{align*}
r_{(M/e)^*}(A)=|A|-r_{M/e}(E_0)+r_{M/e}(E_0\setminus A).
\end{align*}
By the contraction rank formula in $M$,
\begin{align*}
r_{M/e}(E_0)=r_M(E_0\cup\{e\})-r_M(\{e\})=r_M(E)-r_M(\{e\}).
\end{align*}
Also $E_0\setminus A=E\setminus(A\cup\{e\})$, so
\begin{align*}
(E_0\setminus A)\cup\{e\}=E\setminus A.
\end{align*}
Applying the contraction rank formula again gives
\begin{align*}
r_{M/e}(E_0\setminus A)=r_M(E\setminus A)-r_M(\{e\}).
\end{align*}
Substituting these two expressions into the dual rank formula yields
\begin{align*}
r_{(M/e)^*}(A)=|A|-\bigl(r_M(E)-r_M(\{e\})\bigr)+\bigl(r_M(E\setminus A)-r_M(\{e\})\bigr).
\end{align*}
The two occurrences of $r_M(\{e\})$ cancel, and therefore
\begin{align*}
r_{(M/e)^*}(A)=|A|-r_M(E)+r_M(E\setminus A).
\end{align*}
[/step]
[step:Compare with the deletion of the dual and conclude]
Let $A\subset E_0$ be arbitrary. Since deletion preserves ranks of subsets not containing the deleted element,
\begin{align*}
r_{M^*\setminus e}(A)=r_{M^*}(A).
\end{align*}
Applying the dual rank formula in $M$ to the subset $A\subset E$ gives
\begin{align*}
r_{M^*}(A)=|A|-r_M(E)+r_M(E\setminus A).
\end{align*}
Thus
\begin{align*}
r_{M^*\setminus e}(A)=|A|-r_M(E)+r_M(E\setminus A).
\end{align*}
This agrees with the formula obtained for $r_{(M/e)^*}(A)$, so
\begin{align*}
r_{(M/e)^*}(A)=r_{M^*\setminus e}(A)
\end{align*}
for every $A\subset E_0$. Both matroids have ground set $E_0$, so the identity map
\begin{align*}
\operatorname{id}_{E_0}:E_0\to E_0
\end{align*}
is a matroid isomorphism. Therefore
\begin{align*}
(M/e)^*\cong M^*\setminus e.
\end{align*}
Together with the first isomorphism, this proves that duality interchanges deletion and contraction.
[/step]
