Assume that $f: \mathbb{R} \to \mathbb{R}$ is a deterministic, increasing function (not necessarily strictly increasing).
Also, assume that $\epsilon_1$ and $\epsilon_2$ are iid noise; i.e., for a distribution $g$, $\epsilon_1 \sim g$ and $\epsilon_2 \sim g$, and $\epsilon_1$ and $\epsilon_2$ are independent.
We know that for known constants $a,b\in \mathbb{R}$, such that $a>b$, we have $$P_{g}(f(a+\epsilon_1)>f(b+\epsilon_2)) \geq P_{g}(f(a+\epsilon_1)<f(b+\epsilon_2)).$$
Now, if I know $$E_{g}[f(a+\epsilon_1)]>E_{g}[f(b+\epsilon_2)],$$ can I conclude: $$P_{g}(f(a+\epsilon_1)>f(b+\epsilon_2)) > P_{g}(f(a+\epsilon_1)<f(b+\epsilon_2)).$$
Assuming that $$ P\big(f(a+\epsilon_1)>f(b+\epsilon_2)\big) = P\big(f(a+\epsilon_1)<f(b+\epsilon_2)\big), $$ we have, taking complements and using the symmetry, $$ P\big(f(a+\epsilon_1)\ge f(b+\epsilon_2)\big) = P\big(f(b+\epsilon_1)\ge f(a+\epsilon_2)\big). $$ However (as you have pointed out), $$ P\big(f(a+\epsilon_1)\ge f(b+\epsilon_2)\big) \ge P\big(f(b+\epsilon_1)\ge f(b+\epsilon_2)\big)\ge P\big(f(b+\epsilon_1)\ge f(a+\epsilon_2)\big), $$ so the left inequality must be equality. Therefore, $$ P\big(f(a+\epsilon_1)\neq f(b+\epsilon_1), f(b+\epsilon_1)\ge f(b+\epsilon_2)\big) = 0. $$ This yields, in view of independence, $$ P\big(f(a+\epsilon_1)\neq f(b+\epsilon_1), f(b+\epsilon_1)\ge f(b+x)\big) = 0 $$ for almost all $x\pmod{g}$. Letting $x\to \operatorname{essinf} \epsilon_1$, we get $$ P\big(f(a+\epsilon_1)= f(b+\epsilon_1)\big) = 1. $$ Consequently, $$ E\big(f(a+\epsilon_1)\big)= E\big(f(b+\epsilon_1)\big) = E\big(f(b+\epsilon_2)\big), $$ a contradiction.
I believe there is a simpler argument (because the statement is quite obvious), but I do not see it.