What is $V_{ce}$ exactly? Is it the sum of $V_{eb}$ and $V_{be}$? Can it be freely controlled assuming control of $V_{cc}$ and $V_b$?

$V_{ce}$ is just the voltage difference between the collector and emitter, i.e., $V_{ce}=V_c - V_e$.

Well, $V_{be}$ and $V_{eb}$ are the same voltage. But indeed, voltage drops must sum up to a total, by Kirchoff's Loop Rule: referring to the transistor switch in Eggleston Fig 4.5, you can write $V_{cc}-I_c R_c - V_{ce}=0$.

$V_{ce}$ can also be written as $V_{ce}= V_{cb}+V_{be}$ (maybe this is what you meant?), although this is not especially useful for analyzing this switch.

Yes, could say that $V_{ce}$ can be controlled for this switch by controlling the voltage at the base and $V_{cc}$ (and $R_c$). By choosing $V_1$, you effectively set $I_b$ and $V_{be}$ (according to the graphical solution depicted in Eggleston Fig. 4.6). This $I_b$ then determines which curve from Fig. 4.7 applies. Along with the choice of $V_{cc}$, this determines both $I_c$ and $V_{ce}$, which can be found from the graphical solution depicted in Fig. 4.7.

(Note that so long as $V_1$, and hence $I_b$, are large enough, the values of $V_{cc}$ and $I_c$ will be in the saturation parts of the curves in Fig. 4.7, and so will be only weakly dependent on the exact value of $V_1$. $V_{ce}$ will be small, and $I_c$ large, where all the curves scrunch together.)