The following solution is due to Brian Lawrence. We will prove the result with the generality of any measurable set $\mathcal{P}$ (rather than a polygon). For a vector $v$ in the plane, write $\mathcal{P} + v$ for the translate of $\mathcal{P}$ by $v$.
Suppose $\mathcal{P}$ is a polygon of area 1, and $\varepsilon > 0$ is a constant, such that for any translate $Q = \mathcal{P} + v$, where $v$ has length exactly $\frac{1}{100}$, the intersection of $\mathcal{P}$ and $Q$ has area at least $1 - \varepsilon$. The problem asks us to prove a lower bound on $\varepsilon$.

Lemma
Fix a sequence of $n$ vectors $v_1, v_2, \dots, v_n$, each of length $\frac{1}{100}$. A grasshopper starts at a random point $x$ of $\mathcal{P}$, and makes $n$ jumps to $x + v_1 + \dots + v_n$. Then it remains in $\mathcal{P}$ with probability at least $1 - n\varepsilon$.
Proof. In order for the grasshopper to leave $\mathcal{P}$ at step $i$, the grasshopper's position before step $i$ must be inside the difference set $\mathcal{P} \setminus (\mathcal{P} - v_i)$. Since this difference set has area at most $\varepsilon$, the probability the grasshopper leaves $\mathcal{P}$ at step $i$ is at most $\varepsilon$. Summing over the $n$ steps, the probability that the grasshopper ever manages to leave $\mathcal{P}$ is at most $n\varepsilon$. $\square$

Corollary
Fix a vector $w$ of length at most 8. A grasshopper starts at a random point $x$ of $\mathcal{P}$, and jumps to $x + w$. Then it remains in $\mathcal{P}$ with probability at least $1 - 800\varepsilon$.
Proof. Apply the previous lemma with 800 jumps. Any vector $w$ of length at most 8 can be written as $w = v_1 + v_2 + \dots + v_{800}$, where each $v_i$ has length exactly $\frac{1}{100}$. $\square$

Now consider the process where we select a random starting point $x \in \mathcal{P}$ for our grasshopper, and a random vector $w$ of length at most 8 (sampled uniformly from the closed disk of radius 8). Let $q$ denote the probability of staying inside $\mathcal{P}$ we will bound $q$ from above and below.
* On the one hand, suppose we pick $w$ first. By the previous corollary, $q \ge 1 - 800\varepsilon$ (irrespective of the chosen $w$).
* On the other hand, suppose we pick $x$ first. Then the possible landing points $x + w$ are uniformly distributed over a closed disk of radius 8, which has area $64\pi$. The probability of landing in $\mathcal{P}$ is certainly at most $\frac{[\mathcal{P}]}{64\pi}$.
Consequently, we deduce
$$1 - 800\varepsilon \le q \le \frac{[\mathcal{P}]}{64\pi} \implies \varepsilon > \frac{1 - \frac{[\mathcal{P}]}{64\pi}}{800} > 0.001$$
as desired.