Let $\{X(t); t\geq 0\}$ be a Poisson process of rate $\lambda$ that represents the arrival process of customers entering a store. I call $W_1,W_2,\dotsc$ the arrival time of the $k$th customer, and $Y_1,Y_2,\dotsc$ the amount of time that the customer spends in the store. Each $Y_k$ are independent random variables having a common distribution function $$G(y) = P(Y_k \leq y).$$ Finally, let $M(t)$ denote the number of people still in the store at time $t$, and $N(t)$ denote the number of people that arrived and departed by time $t$. The problem is to determine the joint distribution of $M(t)$ and $N(t)$.
So what I have is, \begin{align*} P(M(t) = j, N(t) = k|X(t) = n)&= P\left[\sum_{i=1}^n\boldsymbol 1\left\{W_i+Y_i \geq t\right\}=j,\right.\\ &\qquad+ \left.\sum_{i=1}^n\boldsymbol 1\{ W_i+Y_i<t = k|X(t) = n\right]\\ &=P\left[\sum_{i=1}^n \boldsymbol 1\{U_i+Y_i \geq t\} = j, \sum_{i=1}^n\boldsymbol 1\{U_i+Y_i<t\} = k\right] \end{align*} where the $U_i$ are independent $unif(0,t)$ because of symmetry and a theorem. But now I don't know where to take this from here. I'm not even sure that I am on the right track. I was hoping for some help and feedback. Thanks!
I think you have the right idea. Since $X(t)=M(t)+N(t)$, conditioning on $X$ gives us
$$P(M(t)=m,\; N(t)=n) = P(X(t)=m+n)P(M(t)=m,\; N(t)=n\mid X(t)=m+n).$$
Firstly, $P(X(t)=m+n) = \dfrac{(\lambda t)^{m+n}}{(m+n)!}e^{-\lambda t}$.
Next, given $X(t)=m+n$, the arrival time of each of those $m+n$ customers is uniformly distributed on $(0,t)$. Further, to get $M(t)=m$ and $N(t)=n$, we need any $n$ of those $m+n$ customers to have left the store by time $t$. So, given $X(t)=m+n$, we have $N(t)\sim Bin(m+n,p)$ where
\begin{align} p &= P(\text{Customer, having arrived by time $t$, leaves by time $t$}) \\ &= \int_{u=0}^{t} P(\text{Cust arrives at time $u$})P(\text{Customer spends less than $t-u$ time in the store}) \\ &= \int_{u=0}^{t} \dfrac{1}{t}G(t-u)\;du \\ &= \dfrac{1}{t} \int_{u=0}^{t} G(u)\;du. \end{align}
\begin{align} \therefore\quad P(M(t)=m,\; N(t)=n\mid X(t)=m+n) &= \binom{m+n}{n}p^n(1-p)^m \\ & \\ \therefore\quad P(M(t)=m,\; N(t)=n) &= \dfrac{(m+n)!}{m!n!}p^n(1-p)^m \dfrac{(\lambda t)^{m+n}}{(m+n)!}e^{-\lambda t} \\ &= \dfrac{(p\lambda t)^n ((1-p)\lambda t)^m}{n!m!} e^{-\lambda t}. \end{align}