I think the following is true, but can't show it.
Let $p$ be a prime number, and let $f(x)$ be a polynomial which satisfies $${\Bigl(1-x+\frac{1}{x}\Bigr)}^p-1=f(x)+f\Bigl(-\frac{1}{x}\Bigr).$$ Then $$f(0)\neq0\pmod{p^2}.$$
This question is equivalent to my previous question.
In this answer I'll prove the identity $$4f(0)\equiv -p\sum_{k=1}^{(p-1)/2}\frac{5^k-1}k\pmod{p^2}\tag 1$$ which can be useful in order to solve the main problem.
If $p\equiv 1\pmod 5$ or $p\equiv 4\pmod 5$, then the equation $x^2-x-1$ has two roots in $\Bbb Z/p\Bbb Z$, say $u,v$. If $\delta$ denote the $p$-derivation $x\mapsto\frac{x-x^p}p$, then from (1) we get: $$4f(0)\equiv-2p(u-3)\delta(u)\pmod{p^2}$$ Since $u\not\equiv 3\pmod p$, we get $f(0)\equiv 0\pmod{p^2}$ if and only if $\delta(u)\equiv 0\pmod p$ which is equivalent to $u^{p-1}\equiv 1\pmod{p^2}$.
proof of (1): Starting from \begin{align} \left(x-\frac 1x+1\right)^p &=\sum_{a+b+c=p}\frac{p!}{a!b!c!}x^a\left(-\frac 1x\right)^b\\ &=\sum_{a+b+c=p}\frac{p!}{a!b!c!}(-1)^bx^{a-b} \end{align} we get for odd $p$ \begin{align} 2f(0) &=-1+\sum_{2a+c=p}\frac{p!}{(a!)^2c!}(-1)^b\\ &=-1+\sum_{a=0}^{(p-1)/2}\frac{p!}{(a!)^2(p-2a)!}(-1)^a\\ &=\sum_{a=1}^{(p-1)/2}\frac{p!}{(a!)^2(p-2a)!}(-1)^a\\ &=p\sum_{a=1}^{(p-1)/2}\frac{(p-1)!}{(a!)^2(p-2a)!}(-1)^a \end{align} Now \begin{align} \frac{(p-1)!}{(p-2a)!} &=\prod_{n=1}^{2a-1}(p-n)\\ &\equiv-\prod_{n=1}^{2a-1}n\\ &\equiv -(2a-1)!\pmod p \end{align} hence $$2f(0)\equiv -p\sum_{a=1}^{(p-1)/2}(-1)^a\frac{(2a-1)!}{(a!)^2}\pmod{p^2}$$ Recall that $$(-1)^a\binom{2a}a=4^a\binom{-1/2}a\equiv 4^a\binom{(p-1)/2}a\pmod p$$ Starting from $$\sum_{a=0}^{(p-1)/2}\binom{(p-1)/2}ax^a=(1+x)^{(p-1)/2}$$ we get \begin{align} \sum_{a=1}^{(p-1)/2}\binom{(p-1)/2}a\frac{x^a}a &=\sum_{a=1}^{(p-1)/2}\binom{(p-1)/2}a\int_0^x t^{a-1}\mathrm dt\\ &=\int_0^x\frac{(1+t)^{(p-1)/2}-1}t\mathrm dt\\ &=\int_1^{1+x}\frac{u^{(p-1)/2}-1}{u-1}\mathrm du\\ &=\sum_{k=1}^{(p-1)/2}\int_1^{1+x}u^{k-1}\mathrm du\\ &=\sum_{k=1}^{(p-1)/2}\frac{(1+x)^k-1}k \end{align} Finally, for $x=4$, we get \begin{align} \sum_{k=1}^{(p-1)/2}\frac{5^k-1}k &=\sum_{a=1}^{(p-1)/2}\binom{(p-1)/2}a\frac{4^a}a\\ &\equiv\sum_{a=1}^{(p-1)/2}\frac{(-1)^a}a\binom{2a}a\\ &\equiv 2\sum_{a=1}^{(p-1)/2}(-1)^a\frac{(2a-1)!}{(a!)^2}\pmod p \end{align} from which the assertion follows.