What is meant by $X\sim \text{Bernoulli}(p)-p$?
This cannot mean that the pdf of $X$ is $f-p$, where $$ f(x) = \left\{ \begin{array}{ll} p & \quad x =1 \\ 1-p & \quad x = 0 \end{array} \right. $$ is the pdf of the Bernoulli distribution since $1-2p$ can be negative.
Excerpt: (From page numbered 25 of http://www.stats.ox.ac.uk/~reinert/talks/steinintroduction.pdf) If $X\sim \text{Bernoulli}(p)-p$, then $$\mathbb{E}\lbrace XI(X>x)\rbrace=p(1-p)$$ for $-p<x<1-p$ and is zero elsewhere (where $I$ is the indicator function).
Bernoulli($p$) yields 1 with probability $p$ and 0 with probability $1-p$.
Bernouilli($p$)$-p$ yields $1-p$ with probability $p$ and $-p$ with probability $1-p$. So in other word $X+p \sim \text{Bernoulli}(p)$.
So then, \begin{equation} E[X I_{\{X>x\}}] = (1-p)\times P(X=1-p) + 0 \times P(X=-p) = p(1-p) \end{equation} for $-p<x<1-p$.
This follows because $X I_{\{X>x\}}$ is $1-p$ with probability $p$ and $0$ with probability $1-p$