We know in a Poisson process with parameter $\lambda$, the time between two consecutive occurrence is exponentially distributed with parameter $\lambda$, and as a result it is memoryless.
We are interested in finding the density function for the elapsed time. Indeed we are looking at the process at a time, what is the density of the time difference of the previous occurrence and the time that we are looking at the process?
I have the following explanation, but I cannot figure out, where am I wrong.
Let $T_e$ be the elapsed time, $T_w$, the time we should wait for the next occurrence, and $T$ be the time between two occurrence. Then obviously we have $$T_e=T-T_w\Rightarrow \mathbb{E}[T_e]=\mathbb{E}[T]-\mathbb{E}[T_w]=\frac{1}{\lambda}-\frac{1}{\lambda}=0$$
and as $T_e\geq 0$, and it has a zero mean, it means that almost everywhere its density function is zero, so $$f_{T_e}(t)\stackrel{a.e.}{=}0$$ which doesn't make sense to be almost everywhere equal to zero. Is it wrong?
To be clear: $T_e$ is the time since the last occurrence, $T_w$ is the time until the next occurrence, in a Poisson process, and $T$ is the time between the occurrences (thus $T=T_w+T_e$).
It is not obvious why you think the time-displacement, $T$, between the last and next occurrence should be conditionally exponentially distributed when given a fixed time-point that is between the occurrences (the time of observation).
The time between an arbitrary pair of subsequent occurrences is exponentially distributed. However these are not arbitrary occurrences.
From the point of view of the last occurrence, $T$ is the time until the next occurrence given that it is after you arrive. The time until you arrive is exponentially distributed, and the time after that until the next occurrence is also exponentially distributed, so their sum is Gamma distributed.
$$~~~~T\sim\Gamma(2,\lambda^{-1}) \\\therefore~~ \mathsf E(T)= 2\lambda^{-1}$$