Showing a Poisson process is stationary 
Let $\{N_t: t\ge 0\}$ be a Poisson Process with rate $\lambda$ and $T_0$ is an independent r.v. where
$$ \Bbb{P}(T_0=1)=\mathbb{P}(T_0=-1)=\frac12$$
If we then let $T_t = T_0(-1)^{N_t},$ how can we show that $T_t$ is stationary?

To show $T$ is stationary, then $TP=T$ where $P$ is a transition matrix. Then we need to prove that $(I-P)T_t=(I-P)T_0(-1)^{N_t} = 0$.
Firstly, the transition matrix of $N_t$ is calculated. So, $N_t|_{N_s=x} = N_t - N_s + N_s|_{N_s=x} = N_{t-s}+x.$ Hence the transition matrix for $N$ is $P(N_{t-s}=y-x)=\frac{e^{-\lambda(t-s)}\lambda(t-s)^{y-x}}{(y-x)!}$
What would the transition matrix of $T$ then be? I have
\begin{aligned} (I-P)T_t &= (I-P)(1)(-1)^{N_t}+(I-P)(-1)(-1)^{N_t}\\&
=(-1)^{N_t}(I-P + P-I)\\
&= 0 
\end{aligned}
but not sure if this is right?
 A: Let me start by clarifying some of your notation. When you say that the transition matrix for $N$ is given by
$$ P(N_{t-s} = y - x) = \frac{e^{-\lambda(t-s)} [\lambda (t-s)]^{y-x}}{(y-x)!}, $$
this seems to refer to the matrix of transition probabilities $p(x,y) = P(N_t = y \, | \, N_s = x)$ of going to state $y$ at time $t$, given that we start from state $x$ at time $s$. You can form a matrix from these numbers by letting $x$ index the rows and $y$ the columns. This seems to be the matrix you are calling $P$, but notice that the entries of this matrix actually depend on the times $s$ and $t$; it would be better to call this matrix $P_{s,t}$, and to call the entries $p_{s,t}(x,y)$.
In your equation $TP = P$, "$T$" should be a row vector encoding the distribution of $T_0$ (at time zero), and "$P$" should be a matrix encoding the transition probabilities of this process from time zero to time $t$. Notice that this "$T$" is completely different from the process itself; it does not refer to $T_t = T_0 (-1)^ {N_t}$. Really, we should rename these $\mu_0$ and $P_{0,t}$, so we must show that $\mu_0 P_{0,t} = \mu_0$, where
$$ \mu_0 = \begin{bmatrix} P(T_0 = 1) & P(T_0 = -1) \end{bmatrix} $$
and
$$ P_{0,t} = \begin{bmatrix} P(T_t = 1 | T_0 = 1) & P(T_t = -1 | T_0 = 1) \\ P(T_t = 1 | T_0 = -1) & P(T_t = -1 | T_0 -1) \end{bmatrix} .$$
All this equation says is that $\mu_0$, the distribution of $T_0$, is the same as $\mu_0 P_{0,t}$, the distribution of $T_t$. All that remains is to calculate the probabilities involved. We know that $P(T_0 = 1) = P(T_0 = -1) = 1/2$. Looking at entry (1,1) of $P_{0,t}$, we have
$$ P(T_t = 1 | T_0 = 1) = P(N_t \text{ is even} | T_0 = 1) = P(N_t \text{ is even}) ,$$
since $N_t$ is independent of $T_0$. Find the other entries of $P_{0,t}$ using the same reasoning. Now the left entry of the row vector $\mu_0 P_{0,t}$ becomes
$$ \frac{1}{2} P(N_t \text{ is even}) + \frac{1}{2} P(N_t \text{ is odd}) = \frac{1}{2} ,$$
and the right entry is also 1/2, which shows that $\mu_0 P_{0,t} = \mu_0$.
