A quick way to generate 3x3 matrices with determinant equal to 1? Perhaps a formula involving the row number and column number of an element or just some parametric equations for each element. I know that I can just multiply two of these matrices together to get another but I'd rather not do that. Also, I wouldn't want too many zeroes in the matrix (1 or 2 at most).

$$\det\left(\pmatrix{1&a&b\cr 0&1&c\cr 0&0&1\cr}\!\!\!
  \pmatrix{1&0&0\cr d&1&0\cr e&f&1\cr}\right)
  =1\times1=1\ .$$
(Does this count as multiplying two of "these" matrices together? - not sure what you meant by "these" matrices.)  It is an easy way to get (for example) integer matrices with determinant $1$, and a bit of trial and error with $a,b,c,d,e,f$ will give you a small number of zeros.
