Textbooks say removing proton from thiamine's C atom will produce a resonance stabilized carbanion which is hence bioactive. That's OK, but how comes that 1) we abstract the acidic proton with the OH base whose pK is lower that for thiamine, and 2) the remaining anion acts - in bio context - in aqua medium (?), hence will immediately deprotonate water and lose its negative charge.
In the ionized structure shown, which looks like an ylid, the carbanion is formed on an $\mathrm{sp^2}$ orbital and no resonance stabilization of this charge is possible.
Regarding your question about the ionization, remember that biochemical reactions don't always happen in water, they may happen in the active site of enzymes, which have a completely different environment: the dielectric constant is different, the $\mathrm{p}K_\mathrm{a}$ of substances change and the proximity of certain groups stabilize reactive intermediates that would not be formed in water.
For a reaction to happen, it is sufficient if 1 in a 1000 thiamine structures are deprotonated. For a species with a $\mathrm{p}K_\mathrm{a}$ of 18 to be 50% deprotonated, you would need an aqueous solution with a pH of 18, which is not possible as far as I know.
So even if thiamine is present in aqueous solution and not (as mentioned in Rafael L's answer) in the active site of an enzyme, it still could be sufficiently deprotonated to show reactions explained by the deprotonated intermediate.
