Ok, here's my first attempt. Close scrutiny and comments appreciated!
The Two-Sample Hypotheses
If we can frame two-sample one-sided Kolmogorov-Smirnov hypothesis tests, with null and alternate hypotheses along these lines:
$\text{H}_{0}\text{: }F_{Y}\left(t\right) \geq F_{X}\left(t\right)$, and
$\text{H}_{\text{A}}\text{: }F_{Y}\left(t\right) < F_{X}\left(t\right)$, for at least one $t$, where:
the test statistic $D^{-}=\left|\min_{t}\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right|$ corresponds to $\text{H}_0\text{: }F_{Y}\left(t\right) \geq F_{X}\left(t\right)$;
the test statistic $D^{+}=\left|\max_{t}\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right|$ corresponds to $\text{H}_0\text{: }F_{Y}\left(t\right) \leq F_{X}\left(t\right)$; and
$F_{Y}\left(t\right)$ & $F_{X}\left(t\right)$ are the empirical CDFs of samples $Y$ and $X$,
then it should be reasonable to create a general interval hypothesis for an equivalence test along these lines (assuming that the equivalence interval is symmetric for the moment):
$\text{H}^{-}_0\text{: }\left|F_{Y}\left(t\right) - F_{X}\left(t\right)\right| \geq \Delta$, and
$\text{H}^{-}_{\text{A}}\text{: }\left|F_{Y}\left(t\right) - F_{X}\left(t\right)\right| < \Delta$, for at least one $t$.
This would translate to the specific two one-sided "negativist" null hypotheses to test for equivalence (these two hypotheses take the same form, since both $D^{+}$ and $D^{-}$ are strictly non-negative):
$\text{H}^{-}_{01}\text{: }D^{+} \geq \Delta$, or
$\text{H}^{-}_{02}\text{: }D^{-} \geq \Delta$.
Rejecting both $\text{H}^{-}_{01}$ and $\text{H}^{-}_{02}$ would lead one to conclude that $-\Delta < F_{Y}\left(t\right) - F_{X}\left(t\right) < \Delta$. Of course, the equivalence interval need not be symmetric, and $-\Delta$ and $\Delta$ could be replaced with $\Delta_{2}$ (lower) and $\Delta_{1}$ (upper) for the respective one-sided null hypotheses.
The Test Statistics (Updated: Delta is outside the absolute value sign)
The test statistics $D^{+}_{1}$ and $D^{-}_{2}$ (leaving the $n_{Y}$ and $n_{X}$ implicit) correspond to $\text{H}^{-}_{01}$ and $\text{H}^{-}_{02}$, respectively, and are:
$D^{+}_{1} = \Delta - D^{+} = \Delta - \left|\max_{t}\left[\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right]\right|$, and
$D^{-}_{2} = \Delta - D^{-} = \Delta - \left|\min_{t}\left[\left(F_{Y}\left(t\right) - F_{X}\left(t\right)\right)\right]\right|$
The Equivalence/Relevance Threshold
The interval $[-\Delta, \Delta]$—or $[\Delta_{2}, \Delta_{1}]$, if using an asymmetric equivalence interval—is expressed in units of $D^{+}$ and $D^{-}$, or the magnitude of differenced probabilities. As $n_{Y}$ and $n_{X}$ approach infinity, the CDF of $D^{+}$ or $D^{-}$ for $n_{Y},n_{X}$ approaches $0$ for $t\le 0$, and must be $>0$ for $t > 0$:
$$\lim_{n_{Y},n_{X}\to \infty}p^{+} = \text{P}\left(\sqrt{\frac{n_{Y}n_{X}}{n_{Y}+n_{X}}}D^{+} \le t\right) = 1 - e^{-2t^{2}}$$
So it seems to me that the PDF for sample size-scaled $D^{+}$ (or sample size-scaled $D^{-}$) must be $0$ for $t<0$, and must be $>0$ for $t \ge 0$:
$$f(t) = {1 - e^{-2t^{2}}}\frac{d}{dt} = 4te^{-2t^{2}}$$
Glen_b points out that this is a Rayleigh distribution with $\sigma=\frac{1}{2}$. So the large sample quantile function for sample size-scaled $D^{+}$ and $D^{-}$ is:
$$\text{CDF}^{-1} = Q\left(p\right) = \sqrt{\frac{-\ln{\left(1 - p\right)}}{2}}$$
and a liberal choice of $\Delta$ might be the critical value $Q_{\alpha}+\sigma/2 = Q_{\alpha}+\frac{1}{4}$, and a more strict choice the critical value $Q_{\alpha}+\sigma/4=Q_{\alpha}+\frac{1}{8}$.