Let $\mathbb{C}_+$ be the right half-plane defined by \begin{align} \mathbb{C}_+ \stackrel{\text{def}}{=} \{z \in \mathbb{C} \ \colon \Re(z) > 0\}. \end{align}
I am trying to understand why for a function $f$ analytic on $\mathbb{C}_+$, and also continuous and bounded on the closure of $\mathbb{C}_+$, we have \begin{align*} \sup_{z \in \mathbb{\mathbb{C}_+}} |f(z)| = \sup_{y \in \mathbb{R}} |f(iy)|. \end{align*}
I rewrote $f(z)$, for $z \in \mathbb{C}_+$, as \begin{align*} f(z) = \int_{-i\infty}^{+i\infty} k_z(\zeta)f(\zeta)dy, \qquad \text{with } \ k_s(\zeta) \stackrel{\text{def}}{=} \frac{ \Re(z)}{i\pi (\zeta - z)(\zeta + \overline{z})} \end{align*} and then wanted to show that \begin{align*} \sup_{z \in \mathbb{\mathbb{C}_+}} \|k_z\|_{L^1(i\mathbb{R})} < \infty, \end{align*} but I failed to do so. I was wondering if someone could help me with this last part, or explain another way to proceed.
Well first, the statement you ask about in the body of the question cannot be literally true, because a priori there's no such thing as $f(iy)$. (In fact the hypotheses do imply that the boundary values $f(iy)$ exist for almost every $y$, but that's a little deeper than this.)
One wonders how you "rewrote" $f$ as claimed. Anyway, one can get a true statement by adding a hypothesis from the title:
This is a simple Phragmen-Lindelof sort of argument. For $\epsilon>0$ define $$f_\epsilon(z)=\frac{f(z)}{1+\epsilon z}.$$Since $f$ is bounded it follows that $$\lim_{z\to\infty}f_\epsilon(z)=0.$$So $|f_\epsilon|$ achieves a maximum at some point of $\overline{\Bbb C_+}$. The Maximum Modulus Theorem says the maximum cannot happen at an interior point, so for every $z\in\Bbb C_+$ we have $$|f_\epsilon(z)|\le\sup_{y\in\Bbb R}|f_\epsilon(iy)|\le\sup_{y\in\Bbb R}|f(iy)|.$$Let $\epsilon\to0$.