This comes from Rudin's Real and Complex Analysis, in which there is a claim, regarding the integral in the title.
(i) Let $0<r<1,\ T$ be the unit circle in $\mathbb C, \ f\in L^1(T),\ $ and $z=re^{it}$. Then the following fact shows that $\int _T\frac{e^{it}+z}{e^{it}-z}f(t)dt$ is a holomorphic function of $z$:
(ii) Suppose $\mu$ is a complex measure on $X$ and $\phi:X\to \mathbb C.$ If $\phi(X)$ does not intersect the open set $\Omega$ in $\mathbb C$ then $f(z)=\int _X \frac{d\mu (x)}{\phi (x)-z}$ is holomorphic for $z\in \Omega.$ So in this exercise, $X=T=[-\pi,\pi]$
Here is my attempt at showing that (ii) implies (i), which I would like feedback on. Is it correct? If so, is there an easier way to do it? If not, where is the error? I think a direct approach would also work, although I have not written it out.
Take $\gamma (t)=e^{it},$ define $d\mu=\gamma'(t)dt,$ and $d \nu_1=fdt,\ d \nu_2=fd\mu$ where $dt$ is Lebesgue measure on $T$. Then, we have
$\int _T\frac{e^{it}+z}{e^{it}-z}f(t)dt=\int _T\frac{e^{it}}{e^{it}-z}f(t)dt+z\int _T\frac{1}{e^{it}-z}f(t)dt.$
So, with $\phi(t)=\gamma (t)=e^{it},$ the first integral is $\frac{1}{i}\int_T\frac{d\nu_2}{\phi(t)-z}$ and the second is $z\int_T\frac{d\nu_1}{\phi(t)-z},\ $ which proves the claim.
I added labels for reference. You don't say what you're trying to do! I think that you're taking (ii) as given and trying to show how (i) follows; if that's what you're trying to do then yes, what you did is fine.
But you need to change the title! The Poisson kernel is certainly not holomorphic. That $(e^{it}+z)/(e^{it}-z)$ thing is not the Poisson kernel.