Let $T$ be a compact linear operator defined as $$ T\circ u = \int_a^b k(x,y)\,u(y)\,dy, $$ where $k(x,y)\in C([a,b]\times[a,b])$ and $k(x,y)\ge0$ for all $x,y$, and $u\in C([a,b])$.
Suppose that the spectral radius of $T$ is less than 1 so that the equation $f=1+T\circ f$ has a unique solution. Can one conclude that $\|(I-T)^{-1}\|_\infty=\|f\|_\infty$?
If the spectral radius of $T$ is less than $1$, then the unique solution of $(I-T)f = 1$ is $$ f = 1 + T1 + T^{2}1+T^{3}1+\cdots . $$ So suppose that $T1=0$. Then the solution is $f=1$ making it unlikely that $\|(I-T)^{-1}\|_{\infty}=\|1\|_{\infty}=1$.
For example, consider $X=C[-1,1]$ and $$ Tf = \frac{1}{2}\int_{-1}^{1}xf(x)\,dx. $$ Then $T$ maps everything to a constant multiple of $1$ and $T1=0$, which gives $T^{2}=0$ and $$ (I-T)^{-1}=I+T+T^{2}+T^{3}+\cdots = I+T. $$ By design, the function $f=1$ is the unique solution of $(I-T)f=1$ because $Tf=0$. Let $g=x$. Then $(I-T)^{-1}g=(I+T)g=x+1/3$. So $$ \begin{align} \|(I-T)^{-1}\|_{\infty} & \ge \frac{\|(I-T)^{-1}g\|_{\infty}}{\|g\|_{\infty}} \\ & = \frac{\|x+1/3\|_{\infty}}{\|x\|_{\infty}} \\ & =\frac{4/3}{1}=4/3 > \|f\|_{\infty}=1. \end{align} $$
Added: I missed the point that your $k \ge 0$. That does give you what you want because $f \le g$ on $[a,b]$ implies $Tf \le Tg$. In particular, $$ |Tf| \le T|f| \le T(\|f\|1)=\|f\|T1. $$ This also holds for all powers of $T$ because $T^{2}$ will have a positive kernel, etc. Therefore, $$ \begin{align} |(I-T)^{-1}f| & \le |(I+T+T^{2}+T^{3}+\cdots)f| \\ & \le (I+T+T^{2}+\cdots)|f| \\ & \le \|f\|(I+T+T^{2}+\cdots)1 \\ & \le \|f\|(I-T)^{-1}1. \end{align} $$ So you definitely have $\|(I-T)^{-1}f\|\le \|(I-T)^{-1}1\|\|f\|$ which implies $$ \|(I-T)^{-1}\| \le \|(I-T)^{-1}1\|. $$ On the other hand, the opposite inequality holds because $$ \|(I-T)^{-1}1\| \le \|(I-T)^{-1}\|\|1\|=\|(I-T)^{-1}\| $$