With P=Ber(p) and Q=Poiss(p) where $p\in (0,1)$, I try to find the TV Distance T(P,Q).
To find the Total Variance Distance we can use the discrete formula
$$T(P,Q) = 1/2 * \sum_{x\in E} |p_\theta(x) - p_{\theta'(x)}|$$
which I translate to $$ T(P,Q) = 1/2 * \sum_0^\infty |p^x (1-p)^{1-x} - \frac{p^x*e^{-p}}{x!}|$$
I then take the seperate case $x=0$ out of the summation and get $$=1/2 * |(1-p)-e^{-p}| + 1/2 * \sum_1^\infty |p^x (1-p)^{1-x} - \frac{p^x*e^{-p}}{x!}|$$
where due to probability $p\in (0,1)$ the second term is 1-P(x=0), therefore $$=1/2 *( |(1-p)-e^{-p}| + |1- (1-p)+e^{-p}|)$$ $$=1/2 *( |(1-p)-e^{-p}| + |p+e^{-p}|)$$
But it seems I am making a mistake somewhere, as it doesn't seem correct.
You've made a few mistakes here. First, $P(x)=p^x(1-p)^{1-x}$ only for $x=0$ and $x=1$, so this term shouldn't be part of the summation for $x>1$.
Second, you say, "the second term is $1-P(X=0)$". What random variable is $X$ supposed to be? It looks like you're assuming that there's random variable $X$ with: $$P(X=x) = \left|p^x(1-p)^{1-x}-\frac{p^xe^{-p}}{x!}\right|$$ so you can use the identity $\sum_1^\infty P(X=x) = 1 - P(X=0)$, but there is no such random variable, because for this "probability", we have $\sum_0^\infty P(X=x) \neq 1$. This is true even if you fix the first problem and get rid of the erroneous $p^x(1-p)^{1-x}$ term for $x>1$.
That said, your overall approach isn't too bad. Try separating out the two cases where $x=0$ and $x=1$, noting that for $x>1$, only the Poisson probability is left: $$\sum_{x=2}^\infty \left|\frac{p^xe^{-p}}{x!}\right|$$ where the absolute value has no effect, so you now have a summation of Poisson probabilities.