What are the consequences of removing a single byte from a sha256 hash?
Does this bias the hash in any way?
  


We want the avalanche criteria on the output bits, that is a change in the any of input bit must randomly affect half of the output bits. Each bit of the hash function must depend on the input bits; removing one bit doesn't affect the others. 


  
  My assumption is this would decrease the computational power needed to reverse the hash by 1/256th. Is that correct?
  


First of all, hash functions are not really reversible since they are compression functions, that is, they map from a large input space to a shorter space.

$$ H:\{0,1\}^* \rightarrow \{0,1\}^l$$

If we want to talk about collision resistance, see the next answer. 
For generic pre-image search, yes; it will decrease the computational power, as you noted.


  
  My assumption is this would increase the likelihood of a hash collision by 25600%. Is that correct?
  


Collision resistance is measured by the generic birthday attack, that is, $\sqrt {2^l}$, $l$ being the output size of the hash function. SHA256 has $\sqrt {2^{256}} = 2^{128}$ generic birthday attack time. 

In your case we will have $\sqrt {2^{256-8}} = 2^{124}$ as generic birthday attack time. Thus, we have a $2^{4}= 16$ speed-up in the attack time. 

TL;DR: Truncating the hash to 31 bytes will be safe (see also this stack exchange answer



Note 1: Bitcoin miners reached $\approx 2^{92}$ SHA-256 hashes per year in 06 Agust 2019..

Note 2: SHA-224 defined in FIPS180-4 is calculated by truncating the SHA-256 hash value with using different initial constants for domain separation so the value is not the same as the first 28 bytes of the SHA-256 value.