Haskell, 69 68 bytes

g x=any(elem x)$scanr(\_->concat.mapM(\y->[y++y,y++map(0-)y]))[[1]]x

Usage example: g [-1,1] -> False.

Even more inefficient than @flawr's answer. It takes too much time and memory for 4 element lists. To see that the list of OVSF codes (with a lot of duplicates) is actually created, try:

take 10 $ c $ scanr(\_->concat.mapM(\y->[y++y,y++map(0-)y]))[[1]] [1..4]

which returns

[[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1]]

i.e. the list starts with all 16 element lists (4 times concatenated, because of [1..4]), continues with all 8 element lists and so on until it ends with [1].

Edit: @xnor saved a byte. Thanks!

Haskell, 69 68 bytes

g x=any(elem x)$scanr(\_->concat.mapM(\y->[y++y,y++map(0-)y]))[[1]]x

Usage example: g [-1,1] -> False.

Even more inefficient than @flawr's answer. It takes too much time and memory for 4 element lists. To see that the list of OVSF codes (with a lot of duplicates) is actually created, try:

take 10 $ c $ scanr(\_->concat.mapM(\y->[y++y,y++map(0-)y]))[[1]] [1..4]

which returns

[[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1]]

i.e. the list starts with all 16 element lists (4 times concatenated, because of [1..4]), continues with all 8 element lists and so on until it ends with [1].

Edit: @xnor saved a byte. Thanks!

Haskell, 69 bytes

c=concat
g x=elem x$c$scanr(\_->c.mapM(\y->[y++y,y++map(0-)y]))[[1]]x

Usage example: g [-1,1] -> False.

Even more inefficient than @flawr's answer. It takes too much time and memory for 4 element lists. To see that the list of OVSF codes (with a lot of duplicates) is actually created, try:

take 10 $ c $ scanr(\n->c.mapM(\y->[y++y,y++[-e|e<-y]]))[[1]] [1..4]

which returns

[[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1]]

i.e. the list starts with all 16 element lists (4 times concatenated, because of [1..4]), continues with all 8 element lists and so on until it ends with [1].

Haskell, 69 68 bytes

g x=any(elem x)$scanr(\_->concat.mapM(\y->[y++y,y++map(0-)y]))[[1]]x

Usage example: g [-1,1] -> False.

Even more inefficient than @flawr's answer. It takes too much time and memory for 4 element lists. To see that the list of OVSF codes (with a lot of duplicates) is actually created, try:

take 10 $ c $ scanr(\_->concat.mapM(\y->[y++y,y++map(0-)y]))[[1]] [1..4]

which returns

[[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1],
 [1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1],
 [1,-1,-1,1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,1],
 [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
 [1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1]]

i.e. the list starts with all 16 element lists (4 times concatenated, because of [1..4]), continues with all 8 element lists and so on until it ends with [1].

Edit: @xnor saved a byte. Thanks!