Just find the most negative weight and add that to all of the weights?

Edit: Oops, as another commenter points out above, this penalizes paths with more steps. I knew it was too easy. :P

Yes, indeed.

I think you can prove that one can't just map the edge weights (so that edge's new weight only depends on its original weight) to make all weights positive, and also preserve all shortest paths.

You may try something more complex, i.e. where the new weight of an edge would depend on some more edges around it. But then the cost of doing so may be close to Bellman-Ford algorithm itself.

Assuming a directed graph. At the entry to each negative edge collect all the nodes reachable in 1 extra hop from that edge and compute the total cost. All that have a positive weight, add as edges from the source of the negative edge. If they still have a negative weight, expand further. Once you're done, drop all the negative edges. This essentially replaces every negative edge with a set of 'shortcuts' that encode all the places the negative edge could have gotten you, such that the shortcuts have positive cost. Use regular Dijkstra on the augmented graph.

This obviously has a potentially exponential complexity, though it will work if there isn't too much negativity. It should at least be enough to convince you that preprocessing is possible absent the existence of negative cost loops.