OOPS, Step 2 is wrong so far (the constant $c$ depends on $t$!). I'll see if I can fix it.
Assume that $a>b$. Put $t=a-b=1-2b$.
Step 1: $$ \begin{aligned} a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right] \\ &\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right] \\& =1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right] \\ &=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right) \end{aligned} $$ (in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)
Step 2. Recall that for $0<t<1$, one has $\lambda^t=c\int_0^{\infty}z^{-t}(1-e^{-\lambda z})\frac{dz}z$ with some positive $c$. Thus $$ (1-t)^t-(a-t)^t=c\int_0^\infty \left(\frac {e^z}z\right)^t(e^{-az}-e^{-z})\frac{dz}z $$ Since for $Q>1$ the Taylor coefficients of $Q^t$ are all positive, we conclude that the Taylor coefficients of $(1-t)^t-(a-t)^t$ are all positive. In particular, the difference is not less than its second order Taylor polynomial on $(0,a)$, which is $$ -t^2+t\log\frac 1{a}+t^2\left[\frac 1a-\frac{\log^2\frac 1a}2\right]\ge -t^2+t\log\frac 1a+t^2\left[1+\log\frac 1a\right] $$
Step 3: We thus have $$ \begin{aligned} b^{2a}&=b(a-t)^{t}\le b\left[(1-t)^t+t^2-t^2-t(1+t)\log\frac 1a\right] \\ &\le b\left[1-t^2-t(1+t)\log\frac 1a\right] \end{aligned} $$ because $(1-t)^t\le 1-t^2$ by Bernoulli.
It remains to add the results of Step 1 and Step 3.