From my "Ultimate" Anti-Debugging reference (see pferrie.host22.com):
The interrupt 0x2D is a special case. When it is executed, Windows uses the current EIP register value as the exception address, and then it increments by one the EIP register value. However, Windows also examines the value in the EAX register to determine how to adjust the exception address. If the EAX register has the value of 1, 3, or 4 on all versions of Windows, or the value 5 on Windows Vista and later, then Windows will increase by one the exception address. Finally, it issues an EXCEPTION_BREAKPOINT (0x80000003) exception if a debugger is present. The interrupt 0x2D behaviour can cause trouble for debuggers. The problem is that some debuggers might use the EIP register value as the address from which to resume, while other debuggers might use the exception address as the address from which to resume. This can result in a single-byte instruction being skipped, or the execution of a completely different instruction because the first byte is missing. These behaviours can be used to infer the presence of the debugger. The check can be made using this code (identical for 32-bit and 64-bit) to examine either the 32-bit or 64-bit Windows environment:
xor eax, eax ;set Z flag
inc eax ;debugger might skip
So you can see that there's nothing asynchronous happening here. The change occurs immediately when the exception occurs.
As far as why it occurs, the skipped byte is intended to be used to pass one byte of additional information at the time of the exception.