>> At a guess, BITS.M should therefore look like <Field type=c_int, 
>> ofs=0:17, bits=1> instead.
>
> Refined guess:  it should be <Field type=c_short, ofs=2:1, bits=1>.

This refined guess seems reasonable.  Although, bitfield allocation order for GCC is dependent on the target ABI.  What you have above is at least consistent with the System V i386 [1] and x86-64 [2] psABIs.  Not sure about others (other targets and MSVC++ related ones).

I tested the original test case plus the cases listed in the i386 psABI, all which seem to work.  I did notice that this doesn't seem to be right for big-endian machines:

>>> from ctypes import *
>>> class S(BigEndianStructure):
...     _fields_ = [("A", c_int, 17), ("B", c_short, 1)]
... 
>>> class T(LittleEndianStructure):
...     _fields_ = [("A", c_int, 17), ("B", c_short, 1)]
... 
>>> s = S()
>>> s.B = 1
>>> s.B
-1
>>> t = T()
>>> t.B = 1
>>> t.B
0

The current implementation got the expected answer of -1 for 't.B' (although that is actually incorrect anyway because bitfields should never be treated as signed).

So some big-endian tests and some tests that check the values stored in the fields will be useful.

Finally, I think proposed allocation seems correct, but I must admit I am not clever enough to follow why the following part works :-)

+        /* Adjust current bit offset if necessary so that the next field
+           doesn't straddle a multiple of 8*dict->size. */
+        if (*pbitofs && (
+                (*pbitofs + bitsize - 1) % (8*dict->size) !=
+                bitsize + (*pbitofs - 1) % (8*dict->size)))
+            *pbitofs += (8*dict->size) - 1 - (*pbitofs - 1) % (8*dict->size);

[1] http://www.uclibc.org/docs/psABI-i386.pdf
[2] http://www.x86-64.org/documentation/abi.pdf