Why isn't cyclopropane considered unsaturated?
In short, there are two definitions with separate purpose.

IUPAC defines alkanes as


  "Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms."


and notes notes about cycloalkanes, contrasting e.g., to unsaturated cycloalkenes:


  "Saturated monocyclic hydrocarbons (with or without side chains), e.g. cyclobutane. Unsaturated monocyclic hydrocarbons having one endocyclic double or one triple bond are called cycloalkenes and cycloalkynes, respectively. Those having more than one such multiple bond are cycloalkadienes, cycloalkatrienes, etc. The inclusive terms for any cyclic hydrocarbons having any number of such multiple bonds are cyclic olefins or cyclic acetylenes."


Thus, as in the earlier comment by @MaxW, cyclopropane equally is a saturated alkane because each carbon atom is connected with the next carbon atom only by a $\ce{C-C}$ single bond.  Or, as MSU condenses, 


  "Alkanes and cycloalkanes are termed saturated, because they incorporate the maximum number of hydrogens possible without breaking any carbon-carbon bonds."


This set of definitions is not the one serving to determine organic structures from spectra, e.g. in Field's training book, where


  "[...] the molecular formular is reduced to $\ce{C_nH_m}$ and the degree of unsaturation is given by $$\mbox{Degree of Unsaturation} = n - \frac{m}{2} + 1$$ The degree of unsaturation indicates the number of $\pi$ bonds or rings that the compound contains."


(4th edition, p. 4; emphasis added by mine.)