Let $T_n$ be the time of the $n^{th}$ arrival time in a Poisson process N with rate $\lambda$.
I want to define something called an excess lifetime process {E(t), t $\geq$ 0} where E(t) = $T_{N(t)+1}$ - t
E(t) can be understood as the time one must wait from time t until the arrival of the next event.
I have a problem deriving the following cdf for E(t),
P(E(t) > x) = $e^{\lambda(x+t)}$ + $\int_0^t$P(E(t-u)>x)($\lambda$$e^{-\lambda e}$)du
I am attempting to condition the process to $T_1$ so that it makes the equation easier, but I am still lost on how to get the integral part of the probability equation.
How do i derive the cdf of E(t)?
I am not sure if I understand your question correctly. Are you trying to derive that specific equation, or are you trying to derive the probability distribution of delays between events in a Poisson process?
If the latter, then I can copy in mine here:
The Poisson Process
A stochastic process that produces independent spikes at an average rate $\lambda$, can be viewed as there being a probability $p$ every discrete time interval $\Delta t = \frac{1}{n}$ that a spike is created, found as \begin{equation} p = \frac{\lambda}{n}, \end{equation} where $n$ is time resolution, i.e. the number of intervals per unit of time.
If the process starts at time $t_0=0$, then the probability that the first spike is created after a time $t=N\Delta t$ is equal to the probability that it is not created in the $N$ first time intervals. The probability that a spike is not created in any given interval is of course $(1-p)$, so the probability that it does not occur after $N$ intervals is therefore $(1-p)^N$, giving us the expression \begin{equation} \text{Prob}(\text{spike after } t) = (1-p)^N. \end{equation} Substituting $N=\dfrac{t}{\Delta t}=n t$ and $p=\dfrac{\lambda}{n}$ gives us \begin{equation} \text{Prob}(\text{spike after } t) = \Big(1- \frac{\lambda}{n}\Big)^{n t}, \end{equation} and in the limit of perfect time resolution $n \to \infty$ we get \begin{align} \text{Prob}(\text{spike after } t) & = \lim_{n \to \infty} \Big(1 - \frac{\lambda}{n}\Big)^{nt} \end{align} which we see is the definition of the exponential function, i.e. \begin{equation} \boxed{\text{Prob}(\text{spike after } t) = e^{-\lambda t}.} \label{Poisson_exp} \end{equation}
The probability that a spike occurs after a time $t$ is in other words exponentially decreasing as $t$ gets larger.
I hope this answers your question.