If $P(a)=0 \Rightarrow P(a+1)=1$ then $P(x)$ has no repeated roots.

Let us suppose that $P \in \mathbb{C}[x]$ has degree $k$, i.e., it has complex roots $\alpha_1,\ldots,\alpha_k$ with $$ P(x)=c\prod_i(x-\alpha_i) $$ where the $\alpha_i$ are distinct complex and $c\neq 0$. In addition, we know that $$ P(x)-1=c\prod_i(x-\alpha_i-1). $$ In particular, the coefficient of $x^{k-1}$ of $P(x)$ verifies $$ -c\sum_i \alpha_i = - c\sum_i (\alpha_i+1). $$ This is impossible whenever $k\ge 2$.

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Edit: The condition of $\alpha_i$ being distinct is necessary. Indeed, the polynomial $P(x):=x^k$ has all roots real and equal to $0$, and $P(a)=0 \implies P(a+1)=1$ for all $a \in \mathbf{R}$.

You need the degree of $P$ to be $d\ge2$. If so then $$P(x)=(x-a_1)(x-a_2)\ldots(x-a_d)$$ for distinct $a_i$. Then $$P(x)-1=(x-a_1-1)(x-a_2-1)\ldots(x-a_d-1).$$ Can you get a contradiction from these?