The error magnification factor is $e^{42}=1.739274941520501\cdot10^{+18}$. You are allowing an absolute error of $10^{-8}$ in the forward integration. Thus it is not astonishing to get an error n the size of $10^{-8+18}=10^{10}$ in the backward integration.

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In more detail: Your DE has a boundary layer. It has an asymptotic solution at $u=\frac1{21}e^{-t}$ and a spring-like mechanism that forces any other solution to rapidly converge toward this asymptotic solution, halving the distance with time step $0.03$. In a crude Taylor interpretation, close to the asymptote the DE reads as $u(t+\frac1{21})=\frac1{21}e^{-t}\iff u(t)=\frac1{21}e^{-t+\frac1{21}}$, so the exact solution will be above the asymptote if the error tolerances are smaller $10^{-3}$.

As this is a moving equilibrium, the allowed error tolerances will be actually realized, as the step size controller detects the boundary of the stability region via the error estimates. Solutions that start far apart can become close enough in their time evolution that their distance is below the absolute tolerance and then become numerically indistinguishable. 

Integrating backwards it becomes apparently random what path they follow, especially as the time steps in the backward integration are not the same as in the forward iteration. Even in fixed step methods the method step error will be different in the backward iteration, it can even have the opposite direction, increasing the round-trip error instead of (partially) erasing it.

The break-out of the tolerance bubble around the shadow of the asymptote will happen very shortly at the start of the backward integration, the distance again doubling with time step $0.03$ and with only a weak memory of where the forward integration started. If the divergence is upward or downward mainly depends on the numerical method, as said above.