In my edition of the book it reads
$$\lim_{n \to \infty} \sup_{\color{red}{u \leq u_0} } \left| \frac{Y(nu)}{n}-u \right|=0 \quad \text{a.s.}$$
So let's prove this. Fix $\varepsilon>0$. For any $n \in \mathbb{N}$ we have by Etemadi's inequality
$$p_n := \mathbb{P} \left( \sup_{u \leq u_0} \left| \frac{Y(nu)}{n}-u\right| > 3\varepsilon \right) \leq 3\sup_{u \leq u_0} \mathbb{P} \left( \left| \frac{Y(nu)}{n}-u\right|>\varepsilon \right). \tag{1}$$
The idea is to show that $$\sum_{n \in \mathbb{N}} p_n<\infty; \tag{2}$$ the claim then follows from the Borel-Cantelli lemma. In order to prove $(2)$ we note that we can choose a constant $C>0$ such that for any $|\lambda| \leq 1$
$$\mathbb{E}e^{\lambda \tilde{Y}_t} \leq e^{Ct \lambda^2}, \tag{3}$$
where $\tilde{Y}_t :=Y_t-t$ denotes the compensated Poisson process; see the lemma below. The (exponential) Markov inequality and $(1)$ then shows
$$\begin{align*} p_n &\leq 3\sup_{u \leq u_0} \mathbb{P} \bigg( Y(nu)-nu>\varepsilon n\bigg)+3\sup_{u \leq u_0} \mathbb{P} \bigg( -(Y(nu)-nu)>\varepsilon n \bigg) \\ &\leq 3 \sup_{u \leq u_0} \bigg[ \exp \left(\lambda \tilde{Y}(nu)-\varepsilon n \lambda\right)+ \exp \left(-\lambda \tilde{Y}(nu)-\varepsilon n \lambda\right) \bigg]. \end{align*}$$
If we choose $\lambda=\frac{1}{\sqrt{n}}$ and apply $(2)$, then we get
$$p_n \leq 6 \exp \left( C u_0-\varepsilon \sqrt{n} \right).$$
Obviously, this entails $(2)$.
Lemma Let $(Y_t)_{t \geq 0}$ be a Poisson process (with rate $1$) and $\tilde{Y}:=Y_t-t$ the compensated Poisson process. Then $(3)$ holds.
Proof: Since $Y_t \sim \text{Poi}(t)$, the exponential moments can be calculated explicitely: $$\mathbb{E}e^{\lambda Y_t} = e^{t \cdot (e^{\lambda}-1)}.$$ Hence, $$\mathbb{E}e^{\lambda \tilde{Y}_t} = e^{t \cdot (e^{\lambda}-1-\lambda)}.$$ For $\lambda \in [-1,1]$, we have $$|e^{\lambda}-1-\lambda| \leq C \cdot \lambda^2$$ and this proves $(3)$.
Remark The claim holds for any Lévy process $(Y_t)_{t \geq 0}$ with finite exponential moments:
$$\lim_{n \to \infty} \sup_{u \leq u_0} \left| \frac{Y(nu)}{n}-\mathbb{E}Y_1 \cdot u \right|=0 \quad \text{a.s.}$$
The answer is negative, moreover, even the convergence in probability may fail.
For example, let $X_{n,i} = n$ with probability $1/n$ and $0$ with probability $1-1/n$. Then, $\mathrm{E}[X_{n,i}] = 1$ for all $n,i$, but
$$
\mathrm P\Bigl(\frac 1n \sum_{i=1}^n X_{n,i} =0 \Bigr) = \Bigl(1-\frac1n\Bigr)^n \to \frac1e,\quad n\to\infty.
$$
(Actually, by the Poisson limit theorem, $\frac 1n \sum_{i=1}^n X_{n,i}$ converges in law to the Poisson distribution with parameter $1$.)
Concerning some positive answers under additional assumptions, see e.g. this topic.
Best Answer
Probably an application of Glivenko-Cantelli's theorem as I suggested in the comments will work.
Assume that the $\xi_t$ are defined on a probability space $(\Omega,\mathcal F,\mu)$. We can assume that $G$ is the cumulative distribution function of a real valued random variable. Let $(\eta_t,t\in\Bbb Z)$ be a collection of independent random variables with cumulative distribution function $G$, defined on $(\Omega',\mathcal G,\nu)$. Using Glivenko-Cantelli's theorem for the product space and the family of i.i.d. random variables $(\xi_t-\eta_t,t\in\Bbb Z)$ gives $$\sup_{u\in\Bbb R}\left|\frac 1n\sum_{i=1}^n\chi_{\{\eta_i(\omega')-\xi_i(\omega)\leqslant u\}}-\mu\otimes\nu\{\eta_i(\omega')-\xi_i(\omega)\leqslant u\}\right|\to 0$$ for $\mu\otimes\nu$ almost every $(\omega,\omega')$. Then integrate with respect to $\nu$.
An alternative approach would use the result of the paper
Ramon van Handel, The universal Glivenko-Cantelli property, Probab. Th. Rel. Fields 155, 911-934 (2013)
which is available here. We consider $\mathcal F:=\{t\mapsto G(x+t), x\in\Bbb R)\}$, which is a Glivenko-Cantelli class as $G$ is bounded. But it seems quite overkill for this problem, since their result is more general.