As pointed out in the comments, the functor $A_*$ cannot in general be lax symmetric monoidal without making some alterations.
Here is an incomplete discussion of when $A_*$ can be lax monoidal.
The first observation is that, for any homotopy associative ring spectrum $A$, the functor $A_*$ naturally takes values in $\pi_* A$-bimodules. The left and right actions of $A$ on itself produce natural left and right actions of $A$ on $A \otimes X$ for any $X$, and lax monoidality of $\pi_*$ show that $A_* X$ is then a $\pi_* A$-bimodule.
The second observation is that this makes the functor $A_*$ a lax monoidal functor to the category of $\pi_* A$-bimodules. The homotopy associativity of $A$ ensures that the two composites $A \otimes A \otimes A \to A \otimes A \to A$ are homotopic, and so for any $X$ and $Y$ the two composites $$ (A \otimes X) \otimes A \otimes (A \otimes Y) \to (A \otimes X) \otimes (A \otimes Y) \to A \otimes (X \otimes Y) $$ are homotopic. The lax monoidality of $\pi_*$ then tells us that the two composites $$ A_* X \otimes \pi_* A \otimes A_* Y \to A_*(X \otimes Y) $$ are equal, establishing $\pi_* A$-bilinearity. However, tracking which maps actually appear on the spectrum level, this specifically uses the right $\pi_* A$-module structure on $A_* X$ and the left $\pi_* A$-module structure on $A_* Y$.
We now assume that the coefficient ring $\pi_* A$ is graded-commutative. We would like to show that, for any $X$, the left and right module structures "coincide": $a \cdot x = \pm x \cdot a$ for any $a \in \pi_* A$ and $x \in A_* X$ according to the Koszul sign rule. If we can do this, then bimodule bilinearity collapses to module bilinearity.
Suppose $x \in A_d X$ comes from a map $S^d \to A \otimes X$. We can express $A$ as a filtered hocolim of finite spectra $A_i$, and get a lift to a map $S^d \to A_i \otimes X$, with an adjoint map $DA_i \otimes S^d \to X$ using Spanier-Whitehead duality. This gives us a lift of $x$ to a factorization $$ S^d \to \xrightarrow{\eta} A \otimes (DA_i \otimes S^d) \to A \otimes X. $$ Therefore, it suffices to check this in the "universal" cases where $x \in A_* (DA_i)$ comes from the canonical map $\eta: S^0 \to A \otimes DA_i$.
This lifts to the limit over $i$, however: the canonical unit map $S^0 \to map(A,A)$ to the function spectrum. In these terms, we are asking if the canonical element $1 \in [A,A]$ is sent to the same element under post-multiplication by $a$ on either the left or the right; or equivalently, if "multiply by $a$ on the left" and "multiply by $a$ on the right" are always homotopic maps $A \to A$, rather than merely giving equal maps $\pi_* A \to \pi_* A$.
I have not managed to sort out the truth or falsehoodAn example of this in general due to timecan be constructed from the following differential graded algebra. Start with the discrete ring $\Bbb Z[x]$. We first form a bimodule with a single generator $y$, but am almost certain that there is anannihilated by $A$ for which$x$ on the left and right; we form a second bimodule with a single generator $z$, annihilated by $x^2$ on the left and $x$ on the right (so it failshas a basis $\{z,xz\}$). An easy way to find one wouldForm the mapping cone $M$ of the map $z \mapsto 2y$: it has $H_0 = \Bbb Z/2$, generated by $y$, and $H_1 = \Bbb Z$, generated by $xz$. Let $A$ be one where the square-zero extension $\Bbb Z[x] \times M$ by this differential graded bimodule $M$. The coefficient ring $A_*$ is graded-commutative but $A_*$ is not centralbecause almost all products are zero. However, in $E_* A$ for some homotopymod-commutative$2$ homology we see the element $E$$z$ which no longer commutes with the generator $x$ from $A_*$.