Take the Zermelo ordinals to be defined by $Z(0) = 0$, $Z(\alpha+1) = \{Z(\alpha)\}$, and $Z(\lambda) = \{Z(\alpha): \alpha<\lambda\}$ (where $\alpha, \lambda$ are Vonvon Neumann ordinals). Then if we add $Z(\omega+ \omega)$ to $V_{\omega +\omega}$ and close under pairing, union, subsets, and powersets, we get a model of $T$ - Choice + $A$ which is not a model of $A^*$.
More precisely, let $D_0 = V_{\omega+\omega} \cup\{Z(\omega+\omega)\}$ and $D_{n+1}$ be the result of adding pairs, unions, subsets, and powersets of element of $D_n$ to $D_n$. Clearly, $M = \bigcup_{n <\omega}D_n$ is transitive and models $T$ - Choice. A simple induction shows that:
For every $x\in D_n$ there is an $\alpha<\omega+\omega$ such that the (Vonvon Neumann) ordinals in $tc(x)$ are less than $\alpha$.
Since $V_{\omega+\omega}\subseteq M$, it follows that the Vonvon Neumann ordinals in $M$ are just those in $\omega+\omega$. So it models $A$. Since the rank of $Z(\omega+\omega)$ is $\omega+\omega$, it doesn't model $A^*$.
(To get Choice in the form ``every set is well-orderable" we just throw in $x \times x$ at $D_{n+1}$ for $x\in D_n$).