For an M/M/1 queueing system, distribution of the time ($t_e$) between two consecutive events (either arrival or departure) can be derived as follows with the independent assumption,
$$F(t_e\ge t)=F(t_a\ge t)\cdot F(t_d\ge t)=e^{-\lambda t}e^{-\mu t}=e^{-(\lambda+\mu)t}$$
where $t_a$ and $t_d$ are the time between two consecutive arrivals and departures respectively.
From the complementary CDF (CCDF) derived above, we can say that $t_e$ is also exponentially distributed (with mean $1/(\lambda+\mu)$).
However, if we derive the result using CDF rather than CCDF, the outcome is,
$$F(t_e\le t)=F(t_a\le t)\cdot F(t_d\le t)\\=(1-e^{-\lambda t})(1-e^{-\mu t})=1-e^{-\lambda t}-e^{-\mu t}+e^{-(\lambda+\mu)t}\\<1-e^{-(\lambda+\mu)t}=1-F(t_e\ge t)$$
From this point of view, it seems we cannot say $t_e$ is exponentially distributed.
I am quite puzzled about this, could anyone shed some light upon this? Thanks.
Your first attempt uses the claim that $$ \{ t_e \ge t \} = \{ t_a \ge t \} \cap \{ t_d \ge t \}. \tag{$\dagger$} $$ (Here the "$\{...\}$" notation just means "the event that [...] happens".)
In your second attempt, you look at $\{t_e \le t\}$. Since everything is with exponential times, "$\le$" or "<" doesn't matter; let's look at $\{t_e < t\}$. You then claim that $$ \{ t_e < t \} = \{ t_a < t \} \cap \{ t_d < t \}. \tag{$\ddagger$}$$
However, these two displays, $(\dagger, \ddagger)$, are incompatible. Note that $\{t_e \ge t\} = \{t_e < t\}^c$. (Here $\{...\}^c$ means the complementary event.) But, for general sets/events $A$ and $B$, $$ \text{if}\quad C = A \cap B \quad\text{then}\quad C^c = A^c \cup B^c. $$ Hence, applying this in your situation, assuming that $(\dagger)$ is the correct statement, we get $$ \{t_e < t\} = \{t_e \ge t\}^c = \bigl( \{t_a \ge t\} \cap \{t_d \ge t\} \bigr)^c = \{t_a < t\} \cup \{t_d < t\}, $$ which you then need to deal with -- basically, you deal with this by doing your first computation for the different parts, and using the independence of $\{t_a \ge t\}$ and $\{t_d \ge t\}$ (which you assumed in your first calculation).