Apart from the answers with iterator-Value from find() and comparison to .end(), there is another way: map::count.
You can call map::count(key) with a specific key; it will return how many entries exist for the given key. For maps with unique keys, the result will be either 0 or 1. Since multimap exists as well with the same interface, better compare with != 0 for existence to be on the safe side.
for your example, that's
return (cars.count(name)>0);
The advantages I see are
1. shorter code,
2. benefit from whatever optimisations the library may apply internally, using its representation details.