In a "purely digital" link where you set an output to "high" and an input the other end of a line is read as "high" then the probability error is purely to do with the SNR of the line. What is the probability that a HIGH can be interpreted as a LOW? By introducing a higher level protocol with error detection and correction you effectively negate most of the SNR errors and the question is now "What is the probability that the protocol cannot correct corrupted bits?"
So yes, the CODEC (or protocol) can be used (and is used) to negate the effects of SNR-induced signal corruption.
As for the second part...
If you assume 1 bit of information is transmitted per quantization level, and 1 bit is received per quantization level, then yes, increasing the quantization level will increase the number of bits sent at any one time. However, the SNR of the transmission medium will then have a greater effect on those now smaller quantization steps, so although you reduce the quantization noise, you now increase the SNR noise.
However, if you don't assume 1 bit per quantization level, but have multiple quantization levels per bit, then you can increase the number of quantization levels and keep the overall bitrate the same, but have more detail about each bit, so can make a better informed decision about what value that bit is.
For instance, you can think of a simple digital link with 2 states (HIGH and LOW) as a 1-bit quantized system. For simplicity we'll call it 1V for HIGH and 0V for low.
Now, you could then have it that anything received >= 0.5V is a HIGH and anything < 0.5V is a LOW. That's 1 bit quantization. 0.5V would be HIGH, but 0.499999999999V would be LOW. That's an infinitesimally small margin for noise.
However, increase the receiving quantization to 2 bits, say, would give you more detail. It would give you 4 voltage levels to consider - 0V, 0.33V, 0.66V and 1V.
You could now say that anything > 0.66V is a HIGH, and anything less than 0.33V is a LOW. You have now introduced a "noise margin". Anything that falls between those values is discarded as noise. The bitrate remains the same, but the overall SNR has fallen.
Then of course you can add a "schmitt trigger" to it (or software equivalent), whereby you toggle the value depending on a transition. When the input rises above 0.66V you see the value as HIGH, and keep it as HIGH. Only when it then drops down below 0.33V do you then switch it to LOW.
For systems where you have discrete voltage levels you could sample them at a higher resolution, and the line-induced noise would occupy the least significant bits of that sampled value. Discarding the noisy bits down to the resolution of the sent data can then reduce the noise in the system. Also taking multiple samples and averaging them, which in effect cancels the random noise out, (known as "oversampling") can reduce the noise as well.
None of those techniques affect the bitrate as such since you're not adding any extra information to the sent values.
First of all, I like the title of the question, "What makes a circuit Digital" ;).
The world around us is analog so the electronics started with exploring the analog components like the ones mentioned by you resistors, capacitors etc. The need for digital circuits was realized when we started building devices like computers, started storing information digitally and many other things.
If you look closely, digital world is based on voltages similar to the analog world. The only difference and the most advantageous feature of digital is that we have reduced the number of variable to just 0 and 1.
Consider a voltage range 0V to 5V.Lets assume that 0V-2.5V is X and 2.5V to 5V is Y. Here we have grouped the voltages in two blocks but the bottom it remains the same. Sam we did with 0 and 1. To make it convenient for ourselves we thought (just for fun!) let X be 0 and Y be 1 and it all started from there. Whether you get a voltage of 1.1V or 2.4V we will interpret is as a 0. Similarly,you get 3.5V or 4.9V it is a 1 for us. This is biggest advantage of digital circuits called Noise Margin.
Since your question was about how a circuit becomes digital, i don't want to get into the technicality much. but if really want to blame someone for this. get hold of this man Claude Shannon. It is believed that it was his article that led to birth of Digital Revolution.
Best Answer
Spehro's comment is basically the answer. Imagine a link which subtracts a random, varying noise value between 0 and 0.5V from the input. You measure 4.5V at one end, but that could mean either a 4.5V input with no noise or 5V minus noise. You can't tell, and have lost your signal entirely.
With a digital signal varying between 5V and 0V over the same link, you can declare everything over 2.5V to be a '1', and lose no signal.