[proofplan]
We prove both identities by comparing rank functions on the common ground set $E \setminus A$. The rank formula for a dual matroid converts ranks in $(M \setminus A)^*$ and $(M / A)^*$ into ranks in $M$, while the rank formula for contraction converts ranks in $M^*/A$ into ranks in $M^*$. After substituting the dual rank formula for $M^*$, the two expressions agree for every subset $X \subset E \setminus A$. Since a matroid is determined by its rank function, the desired matroid identities follow.
[/proofplan]
[step:Fix the common ground set and rank notation]
Let $S := E \setminus A$. For any matroid $N$ on a finite ground set $F$, write $r_N : 2^F \to \mathbb{N} \cup \{0\}$ for its rank function. We use the standard rank formula for the dual matroid: for every $Y \subset F$,
\begin{align*}
r_{N^*}(Y) = |Y| - r_N(F) + r_N(F \setminus Y).
\end{align*}
We also use the standard rank formula for matroid contraction: for every $B \subset F$ and every $Y \subset F \setminus B$,
\begin{align*}
r_{N/B}(Y) = r_N(Y \cup B) - r_N(B).
\end{align*}
We use matroid deletion in the form that, for every $Y \subset F \setminus B$, the deletion $N \setminus B$ has rank $r_{N \setminus B}(Y) = r_N(Y)$.
A matroid is determined by its rank function: a subset $I$ of its ground set is independent exactly when $r(I) = |I|$. Therefore, if two matroids on the same finite ground set have the same rank function, they have the same independent sets and hence are equal as matroids.
[/step]
[step:Compare the rank functions of $(M \setminus A)^*$ and $M^*/A$]
Let $X \subset S$. Since $M \setminus A$ is a matroid on $S$, the dual rank formula gives
\begin{align*}
r_{(M \setminus A)^*}(X) = |X| - r_{M \setminus A}(S) + r_{M \setminus A}(S \setminus X).
\end{align*}
By the definition of matroid deletion, ranks of subsets of $S$ are unchanged from $M$, so
\begin{align*}
r_{(M \setminus A)^*}(X) = |X| - r_M(S) + r_M(S \setminus X).
\end{align*}
On the other hand, by the rank formula for contraction applied to the matroid $M^*$ and the subset $A \subset E$,
\begin{align*}
r_{M^*/A}(X) = r_{M^*}(X \cup A) - r_{M^*}(A).
\end{align*}
Applying the dual rank formula for $M^*$ first to $X \cup A$ and then to $A$, we get
\begin{align*}
r_{M^*}(X \cup A) = |X \cup A| - r_M(E) + r_M(E \setminus (X \cup A)).
\end{align*}
Also,
\begin{align*}
r_{M^*}(A) = |A| - r_M(E) + r_M(E \setminus A).
\end{align*}
Since $X \subset E \setminus A$, the sets $X$ and $A$ are disjoint, so $|X \cup A| - |A| = |X|$. Moreover,
\begin{align*}
E \setminus (X \cup A) = (E \setminus A) \setminus X = S \setminus X.
\end{align*}
Substituting these identities gives
\begin{align*}
r_{M^*/A}(X) = |X| - r_M(S) + r_M(S \setminus X).
\end{align*}
Thus
\begin{align*}
r_{(M \setminus A)^*}(X) = r_{M^*/A}(X)
\end{align*}
for every $X \subset S$.
[guided]
We want to prove equality of two matroids on the same ground set $S = E \setminus A$. The most efficient way is to compare their rank functions on every subset $X \subset S$.
First consider $(M \setminus A)^*$. The matroid $M \setminus A$ has ground set $S$, so the dual rank formula gives
\begin{align*}
r_{(M \setminus A)^*}(X) = |X| - r_{M \setminus A}(S) + r_{M \setminus A}(S \setminus X).
\end{align*}
Matroid deletion does not change the rank of any subset of the remaining ground set. Since both $S$ and $S \setminus X$ are subsets of $E \setminus A$, we have
\begin{align*}
r_{M \setminus A}(S) = r_M(S).
\end{align*}
and
\begin{align*}
r_{M \setminus A}(S \setminus X) = r_M(S \setminus X).
\end{align*}
Therefore
\begin{align*}
r_{(M \setminus A)^*}(X) = |X| - r_M(S) + r_M(S \setminus X).
\end{align*}
Now consider $M^*/A$. Since matroid contraction by $A$ has ground set $E \setminus A = S$, the contraction rank formula gives
\begin{align*}
r_{M^*/A}(X) = r_{M^*}(X \cup A) - r_{M^*}(A).
\end{align*}
The point of using the dual rank formula is that it rewrites both terms entirely in ranks of the original matroid $M$. Applying the dual rank formula to $X \cup A \subset E$ gives
\begin{align*}
r_{M^*}(X \cup A) = |X \cup A| - r_M(E) + r_M(E \setminus (X \cup A)).
\end{align*}
Applying it to $A \subset E$ gives
\begin{align*}
r_{M^*}(A) = |A| - r_M(E) + r_M(E \setminus A).
\end{align*}
Subtracting these expressions, the two terms $r_M(E)$ cancel:
\begin{align*}
r_{M^*/A}(X) = |X \cup A| - |A| + r_M(E \setminus (X \cup A)) - r_M(E \setminus A).
\end{align*}
Because $X \subset E \setminus A$, the sets $X$ and $A$ are disjoint, so
\begin{align*}
|X \cup A| - |A| = |X|.
\end{align*}
Also $S = E \setminus A$, and removing both $X$ and $A$ from $E$ is the same as first passing to $S$ and then removing $X$:
\begin{align*}
E \setminus (X \cup A) = S \setminus X.
\end{align*}
Finally, $E \setminus A = S$, so the preceding expression becomes
\begin{align*}
r_{M^*/A}(X) = |X| - r_M(S) + r_M(S \setminus X).
\end{align*}
This is exactly the formula already obtained for $r_{(M \setminus A)^*}(X)$. Hence the two rank functions agree on every subset $X \subset S$.
[/guided]
[/step]
[step:Conclude the first matroid identity from equality of ranks]
Both $(M \setminus A)^*$ and $M^*/A$ are matroids on the same ground set $S$. The preceding step proves that their rank functions agree on every subset $X \subset S$. Since a matroid is determined by its rank function, we obtain
\begin{align*}
(M \setminus A)^* = M^*/A.
\end{align*}
[/step]
[step:Compare the rank functions of $(M/A)^*$ and $M^* \setminus A$]
Let $X \subset S$. Since $M/A$ is a matroid on $S$, the dual rank formula gives
\begin{align*}
r_{(M/A)^*}(X) = |X| - r_{M/A}(S) + r_{M/A}(S \setminus X).
\end{align*}
Using the contraction rank formula in $M$, we have
\begin{align*}
r_{M/A}(S) = r_M(S \cup A) - r_M(A).
\end{align*}
Since $S \cup A = E$, this is
\begin{align*}
r_{M/A}(S) = r_M(E) - r_M(A).
\end{align*}
Similarly,
\begin{align*}
r_{M/A}(S \setminus X) = r_M((S \setminus X) \cup A) - r_M(A).
\end{align*}
Because $X \subset S = E \setminus A$, we have
\begin{align*}
(S \setminus X) \cup A = E \setminus X.
\end{align*}
Therefore
\begin{align*}
r_{M/A}(S \setminus X) = r_M(E \setminus X) - r_M(A).
\end{align*}
Substituting into the dual rank formula gives
\begin{align*}
r_{(M/A)^*}(X) = |X| - r_M(E) + r_M(E \setminus X).
\end{align*}
For the matroid deletion $M^* \setminus A$, ranks of subsets of $S$ are unchanged from $M^*$. Thus
\begin{align*}
r_{M^* \setminus A}(X) = r_{M^*}(X).
\end{align*}
Applying the dual rank formula for $M^*$ gives
\begin{align*}
r_{M^* \setminus A}(X) = |X| - r_M(E) + r_M(E \setminus X).
\end{align*}
Hence
\begin{align*}
r_{(M/A)^*}(X) = r_{M^* \setminus A}(X)
\end{align*}
for every $X \subset S$.
[/step]
[step:Conclude the second matroid identity from equality of ranks]
Both $(M/A)^*$ and $M^* \setminus A$ are matroids on the ground set $S = E \setminus A$. The preceding step proves that their rank functions agree on every subset of $S$. Since a matroid is determined by its rank function, we conclude that
\begin{align*}
(M/A)^* = M^* \setminus A.
\end{align*}
This proves both deletion-contraction duality identities.
[/step]
