Denote $Q(x)=P(x)-\frac{1}{100}$ and we obtain $(x+100)Q(x)=xQ(x+1)$. So $Q(0)=0$ and hence $Q(x)=xQ'(x)$. We obtain $(x+100)Q'(x)=(x+1)Q'(x+1)$. So $Q'(-1)=0$ and hence $Q'(x)=(x+1)Q''(x)$. Iterating this argument, we obtain $Q(x)=x(x+1)(x+2)\cdots (x+99)R(x)$. Plugging this into the original equation, we obtain $R(x)=R(x+1)$ and hence $R$ is constant. So $\boxed{P(x)=cx(x+1)(x+2)\cdots (x+99)+\frac{1}{100}}$ which is indeed a solution whatever $c \in \mathbb{R}$.

My solution is similar to the above: First note that that plugging in $x=0$ gives $P(0)=\frac{1}{100}$ now we have if $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ that $a_0=\frac{1}{100}$. Now plugging in $x=-100$ we have $P(-99)=\frac{1}{100}$. So now its obvoius it has something to do with $\frac{1}{100}$. If we continue like this we conclude that for $x=-1,-2,..,-99$ $P(-1)=P(-2)=P(-3)=..=P(-99)=P(0)=\frac{1}{100}$. Now if we note that $P(x)-a_0=a_nx^n+a_{n-1}x^{n-1}+...+a_1x=x(a_nx^{n-1}+a_{n-1}x^{n-2}+...+a_1)$ that is $P(x)-\frac{1}{100}=xQ(x)$ we have : $xQ(x)=0$ for $x=-1,-2,...,-99 $ so $Q(x)=x(x+1)(x+2)...(x+99)T(x)$ and $T(x)=T(x+1)$ $\implies$ $T(x)=c$ for some $c \in \mathbb{R}$ so solution is indeed $P(x)=cx(x+1)(x+2)...(x+99)+\frac{1}{100}$ for every $c \in \mathbb{R}$.

Quite easy