It's true if $A$ is Noetherian.
For any $A$-algebra $C$ and any ideal $J$ in $A$, note that $C/JC$ is isomorphic to the tensor product algebra $C \otimes_A A/J$.
Now for any $n \geq 0$, $A/I^n$ is a module over $A^*$, and the multiplication map $A/I^n \otimes_A A^* \to A/I^n$ is an isomorphism, since $A$ is Noetherian --- see for example Proposition 10.13 of Atiyah and Macdonald's "Introduction to Commutative Algebra".
Let $T = A^* \otimes_A B$. Then we get isomorphisms
$T/I^n T \cong A/I^n \otimes_A (A^* \otimes_A B) \cong (A/I^n \otimes_A A^*) \otimes_A B \cong (A/I^n) \otimes_A B \cong B/I^nB$.
Now pass to the inverse limit to obtain an isomorphism $T^* \stackrel{\cong}{\longrightarrow} B^*$.
I expect that this answer is satisfactory, although it isn't a complete answer. This is really the best result one can hope for.
For a $C^\ast$-algebra $A$ let $Prime(A)$ be the prime ideal space (defined exactly as the primitive ideal space, but with prime (two-sided closed) ideals instead). It is well-known that $Prime(A) = Prim(A)$ when $A$ is separable or when $A$ is Type I.
The following result is due to Proposition 2.16 and 2.17 in [Blanchard, Etienne; Kirchberg, Eberhard Non-simple purely infinite Cā-algebras: the Hausdorff case. J. Funct. Anal. 207 (2004), no. 2, 461ā513.] and builds on Kirchberg's deep work on exact $C^\ast$-algebras:
Theorem: If $A$ and $B$ are $C^\ast$-algebra and at least one of them is exact, then $Prime(A\otimes_{\textrm{min}} B)$ is homeomorphic to $Prime(A) \times Prime(B)$.
There is a canonical map $Prime(A) \times Prime(B) \to Prime(A\otimes_{\mathrm{min}} B)$ given by $(I,J) \mapsto I\otimes_{\mathrm{min}} B + A \otimes_{\mathrm{min}} J$, and this is the homeomorphism in the theorem above.
If $I$ and $J$ are primitive ideals in $A$ and $B$ respectively, then $I\otimes_{\mathrm{min}} B + A\otimes_{\mathrm{min}} J$ is also primitive. Hence we get the following corollary for primitive ideal spaces instead of prime ideal spaces.
Corollary: Let $A$ and $B$ be $C^\ast$-algebras for which all prime ideals are primitive (e.g. separable/Type I/simple), and such that at least one of $A$ and $B$ is exact. Then $Prim(A\otimes_{\textrm{min}} B)$ is homeomorphic to $Prim(A) \times Prim(B)$.
The results fail in general (even when $A$ is simple, non-exact, and $B=\mathcal B(\ell^2(\mathbb N))$), so you can't really hope for anything better.
As commutative $C^\ast$-algebras are exact, the above theorem is applicable if one of the $C^\ast$-algebras is commutative.
Note that one gets $Prim(A\otimes_{\mathrm{min}} B) \cong Prim(A) \times Prim(B)$ (with primitive ideal spaces), if, for instance, $A$ is commutative (hence Type I) and $B$ is separable.
Best Answer
Well, i do not know the answer in general but since you are asking for a reference and if
There are some related results, for example:
Let $B$ an arbitrary algebra over a field.
This is theorem 4.3.2, p. 74, from the book of Drozd-Kirichenko.
(although i suspect this may not be of much use for the particular case you are interested in, as explained in the last paragraph of the OP).
Maybe this reference might also be of some interest to you.