Unfortunately not. Just look at the (confluent) limit case when all $\xi_i=0$: in this case $P_n$ is just the $n$-th Taylor polynomial for $f$. which clearly converges to $f$ not only on $(0,1)$, but on the whole complex plane, but $\xi_i$ are all at the origin.

A less extreme argument is: if we consider the sequence $q_n$ of best polynomial approximants to $f$ on a subinterval $(\alpha, \beta)$ of $(0,1)$, then (i) $q_n$ actually interpolate $f$ on $(\alpha, \beta)$ (but not outside), and (ii) $(f-q_n) \to 0$ in the largest ellipse with foci at $\alpha$ and $\beta$ where $f$ has an analytic continuation. Since $f$ is entire, convergence holds on the whole $\mathbb{C}$, so on $(0,1)$.

EDIT after the comments of Dirk and Qiaochu Yuan:

I don't think that the fact that nodes are different or form a sequence makes any difference. Still, what about this argument: the Cauchy formula for the interpolation error at a point $t$ is
$$
(f-p_n)(t)=\frac{f^{(n+1)}(\theta)}{(n+1)!} \omega_n(t), \quad \omega_n(t)=\prod_{i=1}^{n+1}(t-\xi_i).
$$
For $t\in (0,1)$, we have $\|\omega_n\|_\infty\leq 1$, and $f^{(n+1)}(\theta)\leq e$, REGARDLESS the distribution of $\xi_i$'s on $(0,1)$. In particular, they could live on $(0, 1/2)$ only, etc.