Given a Ferrers diagram, prove that $\det(M)=1$

Here is an observation, which I think gives helpful progress towards a solution. Consider how the matrix for $\lambda$ relates to the matrix for $\lambda'$, where $\lambda'$ is obtained by removing a corner square of $\lambda$ which is not in $M$.

In your example, if you remove the lowest square of the Ferrer's diagram, then it turns out none of the numbers change. More interestingly, consider removing the rightmost square in the second row from the bottom, and computing the new matrix $M'$. The result is

6 3 1 ☐ ☐
3 2 1
1 1 1
☐
☐

Now, how does this new matrix $M'$ relate to the old matrix $M$? Note the $M$' can be obtained from $M$ by a single elementary column operation, namely, subtracting the second column in $M$ from the first. In general, $M'$ will be obtained by $M$ by several row or column operations. Essentially, this is because deleting a box from $\lambda$ removes certain paths, all of which corresponded to paths in a different column. Since these column operations do not change the determinant, $\det M=\det M'$, allowing you to conclude $\det M=1$ by induction on the number of boxes in $\lambda$.

You might have to adjust this argument a little bit when $\lambda$ is a square, so that there are no boxes outside of $M$ to remove, but here you can probably directly prove $\det M=1$, since there is a simple formula for the entries of $M$.