The statement is still false when $P$ is connected and graded. It is even false if we add in addition that $P$ has a unique minimal and maximal element.

**Theorem** For any graded finite poset $P$, let $\rho(P)$ be the average of the ranks of the elements of $P$. Suppose that $Q$ is an upper ideal of $P$ and $\rho(Q) < \rho(P)$. (And, equivalently, $\rho(P) < \rho(P \setminus Q)$.) For all sufficiently large $n$, there is no perfect matching between levels of $P^n$ and $Q^n$.

**Proof** Let $P$ be graded with $p_i$ elements in grade $i$. Let $p(t) = \sum p_i t^i$. So $P^n$ is graded with corresponding polynomial $P(t)^n$. I'll write $P(t)^n = \sum p^n_i t^i$. Define $Q(t)$ similarly and $q^n_i$. Suppose that $p^n_i \leq p^n_{i+1}$ but $q^n_i > q^n_{i+1}$. Then there could be no matching between levels $i$ and $i+1$ of $P^n$. The first inequality would force the matching to be upwards (level $i$ injecting into level $i+1$), but the second inequality would imply that there were not enough elements to match level $i$ of $Q^n$ to.

Now, by a result of Odlyzko and Richmond, for sufficiently large $n$, the coefficients of $P(t)^n$ will be unimodal. Also, by the central limit theorem, the coefficient of $t^k$ in $P(t)^n$ will be much larger when $k$ is near $\rho(P) n$ then anywhere else. Combining these results, the coefficients of $P(t)$ increase up to $ (\rho(P) + o(1))n$ and then decrease down the other side.

Similarly, the coefficients of $Q(t)^n$ increase up to $(\rho(Q)+o(1)) n$ and
then decrease. Since $\rho(Q) < \rho(P)$, this shows that the largest coefficient of $Q(t)^n$ occurs before the largest coefficient of $P(t)^n$ for $n$ large. Thus, for $n$ large, there is no matching at the levels between $(\rho(Q)+o(1)) n$ and $(\rho(P)+o(1)) n$. $\square$

Here is are explicit counterexamples. The poset $P$ has $6$ elements, denoted by the symbols $Q$ and $p$; the poset $Q$ is the $3$ elements labeled $Q$:

Then $P(t) = 3+3t$ and $P(t)^n$ has its largest coefficient at $t^{n/2}$, while $Q(t) = 2+t$ and has its largest coefficient at $t^{n/3}$. So, choosing $i$ between $n/3$ and $n/2$, there is no matching between level $i$ and level $i+1$.

Here is a similar example where $P$ has a maximal and minimal element.

We have $\rho(P) = 3/2=1.5$ and $\rho(Q) = (5/7) 1 + (1/7) 2 + (1/7) 3 = 10/7 \approx 1.4$ as desired.