Can we maximize the log of the odds instead of log of the probability?

Yes, because if you have two different probabilities $p_1$ and $p_2$ with $p_1 < p_2$, the following are all equivalent: begin{align} p_1 & < p_2 \ - p_1 & > - p_2 \ 1 - p_1 & > 1 - p_2 \ frac{1}{1 - p_1} & < frac{1}{1 - p_2} end{align} because applying an increasing function does not reverse the inequality (like $f(x) = x + 1$ from line 2 to 3) while applying a decreasing function (like $g(x) = -x$ from lines 1 to 2 and $h(x) = 1/x$ for positive values of $x$ from lines 3 to 4) does reverse the inequality. Then multiplying the first and last lines gives you $$ frac{p_1}{1 - p_1} < frac{p_2}{1 - p_2}. $$ These are also equivalent to $log(p_1) < log(p_2)$ and $$ log left( frac{p_1}{1 - p_1} right) < log left( frac{p_2}{1 - p_2} right) $$ because $log$ is an increasing function.