I think the answer is: It's always continuous if $A$ is subhomogeneous and never continuous otherwise(min or max).

First notice that if the product map is continuous for min, then it's continuous for max because we can factor the product map as $A\otimes_{max}A\rightarrow A\otimes_{min}A\rightarrow A$ where the first map is the canonical quotient and the second is the product map.

I'm nearly positive (double dual arguments always make me nervous) that if $A$ has an infinite dimensional irreducible representation (or just has irreducible finite dimensional representations of arbitrarily large dimension ) that the product map will not extend to a continuous map on $A\otimes_{max} A$ (and therefore also not for the min tensor product for the reason given in the second paragraph). My reasoning is as follows:

Suppose that $P:A\otimes_{max} A\rightarrow A$ is the continuous product map. Then $P^{**}:(A\otimes_{max} A)^{**}\rightarrow A^{**}$ is w*-continuous. By assumption the von Neumann algebra $(A\otimes 1)^{**}\subseteq (A\otimes_{max} A)^{**}$ has a subalgebra isomorphic to $B(H)$ for $H$ infinite dimensional.

As in my comment, let $e_{ij}$ be a system of matrix units for $B(H).$ For each $i$ get a bounded net $a_{\alpha,i}$ from $A$ converging w* to $e_{1i}.$ Get another bounded net $b_{\alpha,i}$ converging to $e_{i1}.$ Set $x_n=\sum_{i=1}^n e_{1i}\otimes e_{i1}:=w*-lim \sum_{i=1}^n a_{\alpha,i}\otimes b_{\alpha,i}.$ Then $x_n$ is a partial isometry so $||x_n||=1.$ But by weak*-continuity of $P^{**}$ (everything in sight is bounded so multiplication is w*-continuous) we have $P^{**}(x_n)=ne_{11}.$

As for the subhomogeneous case we may, by injectivity of $\otimes_{min}$, suppose that $A$ is actually isomorphic to $C(X)\otimes M_n.$ The claim should now follow since the product map on $M_n$ is continuous.