Suppose we have a sequence of r.v's $\{\tau_n\}_{n \geq 1}$ such that $\Delta \tau_n = \tau_n - \tau_{n - 1}$ are i.i.d. and $\lim_{n \to \infty} \tau_n/n = c$ a.s. for a known $0 < c < \infty$. Suppose further that we have another collection of r.v's $\{L_n\}_{n \geq 1}$ and we're given that $\lim_{n \to \infty} L_{\tau_n}/n = \gamma$ a.s., for a known $0 < \gamma < \infty$. I want to show that $\lim_{n \to \infty} L_n/n = \gamma/c$ a.s.
Here is my attempt:
Theorem: Let $Y_n \to Y$ a.s. and $N_t \to \infty $ a.s., then $Y_{N_t} \to Y$ a.s.
Consider the sequence $s_t = \lfloor t/c \rfloor$. Clearly $s_t \to \infty$ a.s., so $\tau_n/n \to c $ a.s. implies $\tau_{s_t}/s_{t} \to c$ a.s. by the Theorem. From there, we use $c s_t \to t$ to get $\tau_{s_t} \to t$ a.s.
Next, $L_{\tau_n}/n \to \gamma$ a.s. implies $L_{\tau_{s_t}}/s_{t} \to \gamma$ a.s., by the Theorem. Using the result of the previous paragraph for $\tau_{s_t}$, $L_{t}/s_{t} \to \gamma$ a.s. and then dividing both sides by $c$ gives $L_{t}/t \to \gamma/c$ a.s.
What you are trying to prove is false.
Let $\tau_n=2n$; the differences $\tau_n-\tau_{n-1}$ are iid constant $2$. Also, $\tau_n/n\to 2$ almost surely.
Let $L_{n}=n$ when $n$ is even, and $L_n=0$ when $n$ is odd. Then $$ \lim_n L_{\tau_n}/n=2n/n=2, $$ but $\lim_n L_n/n$ does not exist, as $L_n/n$ alternates between $2$ and $0$.