One cannot do much better: let $\mathbb{P}(X = 2/a) = p$ and $\mathbb{P}(X = b) = 1 - p$, then 
$$ \mathbb{E} X = \frac{2p}{a} + (1 - p) b, \qquad \mathbb{E} X^{-1} = \frac{ap}{2} + \frac{1 - p}{b} . $$
Solving $\mathbb{E} X = 1$ and $\mathbb{E} X^{-1} = 2$ for $p$ and $b$, we get
$$ b = \frac{a - 2}{a - 4} > 0, \qquad p = \frac{2 a}{a^2 - 4a + 8} \in [0, 1]$$
whenever $a > 4$, and clearly $p \approx \frac{2}{a}$ as $a \to \infty$.