Does each pixel of the original image has a corresponding coefficient for every scale and every direction calculated by the DWT?
No, by essence, as you guessed correctly. There is a direct counting argument: given a $2N\times 2N$ image, the first level of the  DWT generates  three $N\times N$ wavelet subbands. And through the pigeonhole principle, you cannot have a bijection between a $4N^2$ and a $3N^2$ coefficient set. 

As for the contourlet, you can resort to undecimated wavelet versions, for which several implementations are possible. Some allow more directions than those obtained through traditional tensor product. 

However, I consider that it is misleading to believe that one isolated pixel could be easily followed across scales or directions and mapped to one coefficient: this requires smoothing or directional filters taking into account weighted combinations of the surrounding pixels. In other words, one should not expect arbitrary precision on location AND arbitrary scale or orientation determination.