Python 2, 49 bytes

Takes as input a 2D binary matrix \$ a \$, and its size \$ n \$.

lambda a,n:sorted(map(sum,a+zip(*a)))[-2:]==[1,n]

Try it online!

There may be shorter approaches, but this is what I could find for now. Please let me know if the algorithm is incorrect.

Python 2, 49 bytes

Takes as input a 2D binary matrix \$ a \$, and its size \$ n \$.

lambda a,n:sorted(map(sum,a+zip(*a)))[-2:]==[1,n]

Try it online!

There may be shorter approaches, but this is what I could find for now. Please let me know if the algorithm is incorrect.

Explanation

Since there is exactly one line, there should be exactly one row/column which contains \$ n \$ ones. Furthermore, since every \$ 1 \$ in the matrix exists on that line, there can be no other row/column containing more than one \$ 1 \$. Since if there were another \$ 1 \$ existing somwhere not on that line, it would create a row/column with at least two \$ 1 \$s.

Given this, it suffices to check that the two rows/columns with the largest number of \$ 1 \$s have counts of [1, n].