After discussing this with Tim we came up with the following answer:
The first steifel whiteny class $\omega_1$ of $M$ can be written as the following composition:
$$M \to BO(n) \to BO \to BAut(\mathbb{S}) \to BAut(\mathbb{Z}) \simeq B\mathbb{Z}/2$$
But if $M$ is of type $\ge 2$ then $[M,BO]\simeq [\Sigma^\infty M, bo] \simeq 0$ since $bo$ is of height $\le 1$. So $M$ must be orientable in cotradiction with the third point.
Conclusion: All closed smooth manifolds are of type $\le 1$.
Oh and I believe that at odd primes, type $1$ complexes can be realized by Lens manifolds. Here I was uncareful. This is wrong as it conflicts with the Tim's third point as was pointed out by Gregory Arone in the comments.
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 $\pi_* A \to \pi_* A$.
An example of this can 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 $y$, annihilated by $x$ 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 has a basis $\{z,xz\}$). Form 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 the square-zero extension $\Bbb Z[x] \times M$ by this differential graded bimodule $M$. The coefficient ring is graded-commutative because almost all products are zero. However, in mod-$2$ homology we see the element $z$ which no longer commutes with the generator $x$ from $A_*$.
Best Answer
Yes, this is true, and appears in the early literature, although I do not immediately remember exactly where; I'd guess work of Wilson and/or Ravenel. I'll assume that $h>0$ for notational simplicity. There is a homology theory $E$ with $E_*=\mathbb{F}_p[v_h,v_{h+1}^{\pm 1}]$, and this is a discrete valuation ring in the graded sense. It follows that when $X$ is finite, $E_*(X)$ is a finite sum of $n$ terms like $E_*$ and $m$ terms like $E_*/v_h^k$, for some $n,m\geq 0$. It follows that $v_h^{-1}E_*(X)$ is isomorphic to a direct sum of $n$ copies of $v_h^{-1}E_*$. Using the fact that all formal groups of strict height $h$ become isomorphic after faithfully flat extension, we find that $|X|_h=n$. On the other hand, we have a cofibre sequence $\Sigma^{|v_h|}E\to E\to K(h+1)$, giving a short exact sequence relating $K(h+1)_*(X)$ to the cokernel and kernel of multiplication by $v_h$ on $E_*(X)$. This gives $|X|_{h+1}=n+2m$.