Why is thiourea a monodentate ligand?
This is not really a standalone answer, rather a pointer or an exeption. There are few complexes where deprotonated thiourea $\ce{S=C(NH2)NH}^-$ acts as a bidentate ligand. One example is (aminomethanethioamido)-hydrido-tetrakis(trimethylphosphine)-technetium(III) $\ce{[Tc(P(CH3)3)4(SCN2H3)]+}$ complex[1]:

The fact that thiourea ligand is deprotonated and is coordinated asymmetrically within equatorial plane in $\eta^2-\ce{N,S}$ bonding mode has been proven by NMR and SCXRD studies. From the article [1, p. 4642] (emphasis mine; $\ce{tu}$ = thiourea $\ce{S=C(NH2)2}$):

The $\ce{Tc-N}$ distance, 2.190(13) Å, is long when compared to
known $\ce{Tc(III)}$ complexes. The range for $\ce{Tc-N}$ bonds was found
to be 2.041 Å ... to 2.124 Å ... . This reflects delocalization of negative charge within the bidentate deprotonated thiourea ligand. This elongation is also seen in the $\ce{Tc-S}$ bond, 2.543(4) Å, which is significantly longer than $\ce{Tc-S}$ bonds in the parent compound $\ce{[Tc(\textit{S}-tu)6][Cl]3}$ ... .

As Martin pointed out, the reasons for the observed denticity are not always transparent. Simply put, it seems like in this case in order to be a bidentate ligand, thiourea molecule needs to increase its nucleophilicity and remove steric hindrance caused by $\ce{H}$-atoms lying within the $\sigma_\mathrm{h}$ plane.   Both can be achieved by eliminating a single proton from the thiourea molecule.
Reference

Watson, P. L.; Albanese, J. A.; Calabrese, J. C.; Ovenall, D. W.; Smith, R. G. Inorg. Chem. 1991, 30 (24), 4638–4643. DOI: 10.1021/ic00024a035.