Inverse Mills Ratio for Logit? For $X\sim N(\mu,\sigma^2)$ ,
$$E[X|X>\alpha] = \mu +\sigma \frac{\phi\left(\frac{\alpha-\mu}{\sigma}\right)}{1-\Phi\left(\frac{\alpha-\mu}{\sigma}\right)} $$
Is there an analogous expression for when $X$ is distributed logistically?

You are asking for the conditional expectation
$$    \DeclareMathOperator{\E}{\mathbb{E}}
   \E \left[ X \mid X>\alpha \right]
$$  when $X$ has a logistic distribution. I will only do the case when $X$ has the standard logistic distribution, leaving for you to add location and scale parameters. The density function is then $f(x)=\frac{e^{-x}}{(1+e^{-x})^2}$ and the cdf $F(x)=\frac1{1+e^{-x}}$.
We first find the conditional probability
$$   \DeclareMathOperator{\P}{\mathbb{P}}
   \P\left( X>t \mid X>\alpha \right) =\frac{\P(X>t)}{\P(X>\alpha)}
$$ (for $t>\alpha$).  This expression simplifies to
$$
   \P\left( X>t \mid X>\alpha \right) =\frac{e^{-t}(1+e^{-\alpha})}{e^{-\alpha}(1+e^{-t})}
$$
and by differentiation we find the conditional density function as
$$
f_\alpha(t)= \frac{{\mathrm e}^{-t} \left(1+{\mathrm e}^{-\alpha}\right)}{{\mathrm e}^{-\alpha} \left(1+{\mathrm e}^{-t}\right)}-\frac{\left({\mathrm e}^{-t}\right)^{2} \left(1+{\mathrm e}^{-\alpha}\right)}{{\mathrm e}^{-\alpha} \left(1+{\mathrm e}^{-t}\right)^{2}}
$$
Then, as usual by integration, we find the conditional expectation as
$$
\E \left[ X \mid X>\alpha \right] = \ln \! \left(1+{\mathrm e}^{-\alpha}\right) {\mathrm e}^{\alpha}+\ln \! \left(1+{\mathrm e}^{-\alpha}\right)+\alpha
$$
