The other difference is that *ptr = val; promises the compiler that there will not be any other references to *ptr in existence across this statement, whereas ptr.write(val) makes no such promise.
This doesn't matter so much in this example, but if instead of val, you have foo(ptr), you're at risk of a surprise, since foo might also dereference ptr, and if it does, you now have two references to *ptr in existence at the same time. And you're most likely to get this wrong with static mut, since I can create a new reference to any visible static mut inside foo().
In contrast, if you do ptr.write(foo(ptr)), and foo(ptr) only ever uses ptr.read and ptr.write, you will never have accidental UB from two references existing at once, because you never have one reference, let alone two.