= Solution
A <renewal process> has independent identically distributed, nonnegative, <almost surely> finite inter-arrival times $X_i$, epochs $S_0=0,S_n=\sum_{i=1}^nX_i$, and count $N(t)=\max\{n:S_n\leq t\}$. Allowing zero inter-arrivals is harmless provided $\mathbb P(X_1>0)>0$. The hypothesis $\mathbb E X_1>0$ ensures there is $\varepsilon>0$ with $p=\mathbb P(X_1\geq\varepsilon)>0$.
The <strong law of large numbers> for the indicators gives $S_n\geq\varepsilon\sum_{i=1}^n\mathbf1_{X_i\geq\varepsilon}\to\infty$, so $N(t)$ is finite for finite $t$. Conversely every fixed $S_n$ is finite, so eventually $N(t)\geq n$. On their countable common <probability>-one event, this proves $\boxed{N(t)\to\infty\text{ as }t\to\infty\text{ almost surely}}$.
For fixed $t$, put $L=\lfloor t/\varepsilon\rfloor$. The hinted comparison process has epochs $S_n'=\varepsilon\sum_{i=1}^n\mathbf1_{X_i\geq\varepsilon}\leq S_n$, hence $N(t)\leq N'(t)$. Equivalently,
$$
\mathbb P(N(t)\geq n)\leq\mathbb P(\operatorname{Bin}(n,p)\leq L).
$$
If $p<1$, for $n>L$ this is bounded by $C_{t,p} n^L(1-p)^n$, by summing the first $L+1$ <binomial distribution>. Thus $\sum_ne^{\theta n}\mathbb P(N(t)=n)<\infty$ for $0<\theta<-\log(1-p)$. If $p=1$, $N(t)\leq L$ deterministically. Therefore \b[a strictly positive exponential moment exists] for every fixed finite $t$, with the same allowed $\theta$ for all such $t$. Almost-sure finiteness of the inter-arrivals is implicit in a proper <renewal process>; allowing an infinite inter-arrival with positive <probability> would invalidate the divergence claim.
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