This is not an answer, but a reformulation. Consider all permutations $\sigma$ of the numbers $\{1,2,\dots,2n\}$ such that $$\sigma(1)>\sigma(2),\,\sigma(3)>\sigma(4),\dots,\sigma(2n-1)>\sigma(2n)$$ and also $$\sigma(1)>\sigma(3)>\dots>\sigma(2n-1).$$ Note that these can be identified with perfect matchings on $\{1,2,\dots,2n\}$ ( each $\sigma(2i-1)$ is paired with $\sigma(2i)$). Write these in lexicographic order. For instance, when $2n=6$, there are $15$ such permutations and writing them from largest to smallest gives $654321$, $654231$, $654132$, $645321$, $\dots$, $615243$. Then I think your observation would follow from the statement ``Any two adjacent permutations have distinct parity".
In fact, the pfaffian can be defined as $$\sum_{\sigma}\operatorname{sgn}(\sigma)a_{\sigma(2i-1),\sigma(2i)},$$ summed over such sigma. In your case, this sum will be a sum of $\pm\alpha^{n_{\sigma}}$ for some integers $n_{\sigma}$, and I think it is clear from your condition on $s_i$ that $n_{\sigma}>n_\tau$ if and only if $\sigma>\tau$. So if the statement above holds the signs alternate.
I suggested above that any two adjacent permutations differ by a transposition, but that is certainly false as can be seen from the example 654132,645321 above. But I did check that the claim holds for $2n=6$.