Partial answer
Due to symmetry, assume that $a \ge b$.
By using $a = 1- b$ and $b\in (0, \frac{1}{2}]$, the desired inequality is written as
$$(1-b)^{2b} + b^{2 - 2b} \le (\cos (b - b^2))^{(1-2b)^2}.$$
Let us first prove the case when $b\in (0, \frac{1}{3}]$.
We give the following auxiliary results (Facts 1 through 3). Their proof is not difficult and hence omitted.
Fact 1: $(1-x)^r \le 1 - rx + \frac{1}{2}(r^2-r)x^2$ for $x\in (0, \frac{1}{2}]$ and $0 < r \le 1$.
Fact 2: $\cos y \ge 1 - \frac{1}{2}y^2$ for $y \in (0, +\infty)$.
Fact 3: From Bernoulli inequality, $(1-x)^r \ge 1 - rx$ for $r \ge 1$ and $0 < x < 1$.
From Facts 2 and 3, we have for $b \in (0, \frac{1}{2}]$,
\begin{align}
&(\cos (b - b^2))^{(1-2b)^2} \\
\ge\ & [1 - \tfrac{1}{2}(b - b^2)^2]^{(1-2b)^2}\\
= \ & [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}[1 - \tfrac{1}{2}(b - b^2)^2]^{1+(1-2b)^2}\\
\ge \ & [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}
\Big(1 - \tfrac{1}{2}(b - b^2)^2[1 + (1-2b)^2] \Big).
\end{align}
From Fact 1, by letting $x = b, r = 2b$, we have for $b \in (0, \frac{1}{2}]$,
$$(1-b)^{2b} \le (1-b)(1+b - b^2 - 2b^3).$$
Thus, it suffices to prove that for $b\in (0, \frac{1}{3}]$,
$$(1-b)(1+b - b^2 - 2b^3) + b^{2-2b} \le [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}
\Big(1 - \tfrac{1}{2}(b - b^2)^2[1 + (1-2b)^2] \Big)$$
or
$$\frac{b^2(2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3)}{-b^4+2 b^3-b^2+2}-b^{2-2b}\ge 0$$
or
$$\frac{2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3}{-b^4+2 b^3-b^2+2}-b^{-2b}\ge 0$$
or
$$\ln \frac{2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3}{-b^4+2 b^3-b^2+2} + 2b \ln b \ge 0.$$
It is not difficult with computer. Indeed, denote LHS by $f(b)$. One can prove that $f''(b) > 0 $ for $b\in (0, \frac{1}{3}]$.
Then one can prove that $f'(b) < 0$ for $b\in (0, \frac{1}{3}]$. Omitted.
