Let $X(t)$ be the number of customers in the system at time $t$, $\lambda$ the arrival rate, and $\mu$ the service rate. Assume $\rho:=\frac\lambda\mu<1$. We have the balance equations
$$
\lambda \pi_n = \mu\pi_{n+1},\ n=0,1,\ldots,
$$
which yield the recurrence $\pi_n = \rho^n\pi_0$. From $\sum_{n=0}^\infty\pi_n=1$ we see that $\pi_0=1-\rho$ and hence $\pi_n=\rho^n(1-\rho)$. The expected size in system is
$$
L = \mathbb E[\pi] = \sum_{n=1}^\infty n\pi_n = \sum_{n=1}^\infty n\rho^n(1-\rho) = \frac\rho{1-\rho}.
$$
By Little's law, the mean sojourn time is then
$$
W = \frac L\lambda = \frac1{\mu-\lambda}.
$$
Now, the waiting time is the sojourn time (total time in system) minus the service time:
$$
W_q = W - \frac1\mu = \frac1{\mu-\lambda}-\frac1\mu = \frac\lambda{\mu(\mu-\lambda)} = \frac\rho{\mu-\lambda}.
$$
We can further derive the distribution of the sojourn times. Let $L^a$ be the number of customers in the system immediately before an arrival, and $W_k$ the service time of the $k^{\mathrm{th}}$ customer. It follows that $W = \sum_{k=1}^{L^a+1}W_k$. Conditioning on $L^a$ yields
\begin{align}
\mathbb P(W>t) &= \sum_{n=0}^\infty \mathbb P(W>t\mid L^a=n)\mathbb P(L^a=n)\\
&= \sum_{n=0}^\infty \mathbb P\left(\sum_{k=1}^{L^a+1}W_k>t\mid L^a=n\right)\mathbb P(L^a=n).
\end{align}
By the so-called "Poisson Arrivals See Time Averages" property, we have $\mathbb P(L^a=n)=\pi_n=\rho^n(1-\rho)$, and the sum $\sum_{k=1}^n W_k$ has $\mathrm{Erlang}(n,\mu)$ distribution. So we have
$$
\mathbb P(W>t) = \sum_{n=0}^\infty \sum_{k=0}^n\frac{(\mu t)^k}{k!}e^{-\mu t}\rho^n(1-\rho)
$$
Since the summands are all nonnegative, Tonelli's theorem allows us to interchange the order of summation:
\begin{align}
\mathbb P(W>t) &= \sum_{k=0}^\infty\frac{(\mu t)^k}{k!}e^{-\mu t}(1-\rho)\sum_{n=k}^\infty \rho^n\\
&= \sum_{k=0}^\infty\frac{(\mu t)^k}{k!}e^{-\mu t}\rho^k\\
&= e^{-\mu t}\sum_{k=0}^\infty\frac{(\mu\rho t)^k}{k!}\\
&= e^{-\mu(1-\rho)t}\\
&= e^{-(\mu-\lambda) t}.
\end{align}
So $W$ is exponentially distributed with parameter $\mu-\lambda$. To find the distribution of $W_q$, we condition on $L$ and use the law of total probability:
\begin{align}
\mathbb P(W_q\leqslant t) &= \sum_{n=0}^\infty\mathbb P(W_q\leqslant t, L=n)\\
&= \sum_{n=0}^\infty \mathbb P(W_q\leqslant t\mid L=n)\mathbb P(L=n)\\
&= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+(1-\rho)\cdot\mathsf 1_{\{t=0\}} + \sum_{n=1}^\infty (1-\rho)\rho^n \int_0^t \mu e^{-\mu s}\frac{(\mu s)^{n-1}}{(n-1)!}\\
&= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+\rho(1-\rho)\sum_{n=1}^\infty\rho^n\int_0^t \mu e^{-\mu s}\frac{(\mu\rho s)^{n-1}}{(n-1)!}\ \mathsf ds\\
&= (1-\rho)\cdot\mathsf 1_{\{t=0\}}+\rho(1-\rho)\int_0^t \mu e^{-\mu(1-\rho)s}\ \mathsf ds\\
&= (1-\rho)\cdot\mathsf 1_{\{t=0\}} + 1-\rho e^{-\mu(1-\rho)t)}\cdot\mathsf 1_{(0,\infty)}(t).
\end{align}
