Jelly, 10 bytes
_ƝµṂƇNṖ⁼ṬƇ
Semi-translation of Leo's excellent Husk answer.
_Ɲ Take the vectorized differences between adjacent elements.
(The deltas builtin, I, vectorizes itself instead.)
ṂƇ Filter by minimum, keeping elements containing -1.
N Negate.
Ṗ Remove the last element.
⁼ Is this list equal to
µ the differences
ṬƇ filtered to only elements containing 1?
Ṭ was originally Ṁ (maximum), but this produced a false negative on the first test case, as the largest element of [-1] is -1 rather than 0. Ṭ produces an array with ones at the provided indices and zeros elsewhere, but ignores non-positive indices, producing an empty array (which is falsy) if there are no positive indices.
Jelly, 15 13 bytes
ṛa+ɗ\>TẇM{ʋƝP
-2 removing ṙ1 because the all-zeros pattern can essentially be ignored--if you flip the last bit off, you already know it's the one that's been on the longest
Takes input as an array of Boolean arrays, and outputs 0 or 1.
ɗ\ Cumulatively reduce the input by:
ṛ right argument
a vectorizing-logical-AND
+ the vectorized sum of the arguments.
ṛa+ɗ\ This turns each 1 into how "long" it's been a 1, consecutively.
ʋƝ For each pair of adjacent elements from that result:
T are the indices at which
> the right is less than the left
ẇ a sublist of
M{ the maximal indices of the left?
P Take the product of the results.