Boundedness is enough. To justify the integral representation of $u$, one usually considers the functions $u_r(z)=u(rz)$ with $0<r<1$. These are smooth on the boundary, so the formula is not an issue: $$u_r(z) = \frac{1}{2\pi} \int_0^{2\pi} u_r(e^{i\theta})\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta \tag1$$
As $r\to 1$, we have $u_r(z)\to u(z)$ for every $z$ in the open unit disk $\mathbb D$, because $u$ is continuous there. So the issue is how to pass to the limit in the right hand side of (1). Since we do it with $z$ fixed, the kernel $\frac{1}{2\pi}\frac{1-|z|^2}{|e^{i\theta}-z|^2}$ is a nice bounded function of $\theta$; it creates no problems. (One could even wrap it together with $d\theta$ into a probability measure $\omega_z$ on the boundary, called the harmonic measure with respect to $z$.)
Suppose that as $r\to 1$, $u_r(e^{i\theta})$ converges for almost every $\theta$ to some function, which we can denote $u^*(e^{i\theta})$. (The asterisk is optional but it helps to keep things in order, emphasizing the distinction between a function and its limit values on the boundary.) In the presence of a uniform bound $|u_r|\le M$ the dominated convergence theorem applies; hence, $$\frac{1}{2\pi} \int_0^{2\pi} u_r(e^{i\theta})\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta \to \frac{1}{2\pi} \int_0^{2\pi} u^*(e^{i\theta})\frac{1-|z|^2}{|e^{i\theta}-z|^2}\,d\theta \tag2$$ which answers your question.
Sometimes we don't have a dominating function handy. Instead, it suffices to assume that there exists $p>1$ and $M$ such that $\int |u_r(e^{i\theta})|^p\,d\theta\le M$ for all $r$. Indeed, a bounded sequence in $L^p$ with $1<p<\infty$ has a weakly convergent subsequence. Let $u^*$ be this weak limit. Since the Poisson kernel is a bounded function of $\theta $ (for a fixed $z$), integration against it a continuous linear functional on $L^p$. Thus, (2) holds for a subsequence, which is enough to get the integral representation of $u$. (Using the representation, one can show that the integrals indeed converge as $r\to 1$.)
With $p=1$ the above does not work since $L^1$ is not reflexive. But if $\int |u_r(e^{i\theta})|\,d\theta$ is bounded, then we can find a subsequence that converges in the sense of weak* convergence of measures. Then the integral representation of $u$ involves not the radial limits $u^*$, but some finite signed measure on the boundary, which need not be absolutely continuous.
We have
$$ \frac{1+z}{1-z} = \frac{1+re^{i\theta}}{1-re^{i\theta}} = \frac{1-r^2+2ir\sin{\theta}}{1+r^2-2r\cos{\theta}}, $$
so the function is indeed analytic inside $D(0,1)$.
Your problem is that that the formula
$$ \int_{C(0,r)} P(z) \, dz = 2\pi r P(0) $$
is not true: starting from
$$ \frac{1}{2\pi i}\int_{C(0,r)} \frac{f(z)}{z} \, dz = f(0), $$
which is Cauchy's integral formula (note the difference being the division by $z$), and rewriting it using $dz = ire^{i\theta} d\theta$ gives Gauss's mean value theorem
$$ \frac{1}{2\pi r} \int_0^{2\pi} f(re^{i\theta}) \, d\theta = f(0), $$
and splitting this into real and imaginary parts gives
$$ \frac{1}{2\pi r} \int_0^{2\pi} P(re^{i\theta}) \, d\theta = P(0) $$
and the same with $P \to Q$. This is not the same as $\int_{C(0,r)} P(z) \, dz$, because working backwards, $d\theta=-i dz/z$.
Best Answer
Note that the easiest way to do this type of question is to use either the harmonic representation of $P$ (for term by term integration):
$P(z)=\sum_{n <0}r^{|n|}e^{in\theta}+\sum_{n \ge 0}r^{n}e^{in\theta}=1+\sum_{n \ge 1}(\bar z^n+z^n), z=re^{i\theta}$
or the closed-form of $P$ (for substitution)
$P(z)=\Re \frac{1+z}{1-z}=\frac{1-r^2}{1-2r\cos \theta +r^2}$
Then by definition the first integral is $P(z)-P(ze^{-i\frac{\pi}{2}})=\frac{1-r^2}{1-2r\cos \theta +r^2}-\frac{1-r^2}{1-2r\sin \theta +r^2}$
while the second is:
$\frac{1}{2\pi}\int_{-\pi}^{\pi}(\sum_{n <0}r^{|n|}e^{in\theta}e^{-int}+\sum_{n \ge 0}r^{n}e^{in\theta}e^{-int})(\sum_{-N}^{N}e^{ikt})dt=$
$=\sum_{-N \le n <0}r^{|n|}e^{in\theta}+\sum_{0 \le n \le N}r^{n}e^{in\theta}=1+\sum_{1}^{N}(\bar z^n + z^n)$,
which is the $N$ truncation of it