After this step the $\ce{OH-}$ which was eliminated …
No. $\ce{OH-}$ is never eliminated but formally. To eliminate the oxygen required, it needs to first accept a proton from somewhere leading to $\ce{R-C(O^-)(OH2+)-NHR}$. The hydroxide ion is simply much too bad a leaving group for any $\mathrm{S_N}$ reaction; it must always be transformed into a better one.
This leads to a severe problem in the reaction. Basic conditions hinder the reaction because they decrease the degree of protonation on that oxygen. However, under acidic conditions the amine will be protonated in solution ($\ce{RNH3+}$) and a protonated amine no longer has any lone pairs that can attack nucleophilicly. You’ll want neutral to mildly basic conditions for the reaction to work out.
Another thing to consider: the ‘elimination of a proton’ is a simple acid-base reaction. Proton displacements are typically much more rapid than any carbocation eliminations because protons, being as small as they are, can tunnel. Additionally, you should never compare the stability of a naked proton against that of a naked carbocation in these eliminations. A naked proton will not be generated, it will attach to a base immediately — and there are enough bases around: $\ce{OH-, H2O, RNH2, \dots}$ Furthermore, carbon-heteroatom bonds are pretty stable if the heteroatom side does not equal a good leaving group (e.g. $\ce{NHTs}$). Thus, these are hardly ever cleaved.
In practice, you would not use a particularly acidic or basic medium in this reaction. Rather, you would use acid activation techniques such as the Steglich conditions ($\ce{DCC, DMAP}$).
