Felix is correct in the small $b$ limit. In the very large $b$ limit, on the other hand, the spectra of $J$ and $L$ will coincide, so one needs some more finesse. The following is a rough, approximate attempt at making some intuition.

Consider the case where the first diagonal entry of $J$ is $a_1<0$ with $|a_1|\ll |b|^2$, and all other diagonal entries in $J$ and $L$ are zero. Then you can expand $\det(tI-J)$ along the first row to get $$\det(tI-J)=b^{2n} \left(U_n(t/2)-\frac{a_1 }{b^2} U_{n-1}(t/2)\right),$$
where $U_n$ is the $n$th Chebyshev polynomial of the second kind. For small $a_1$, then, the question is which way does the second term "push" the eigenvalue? I don't have a proof but it's graphically clear that the signs of $U_n$ and $U_{n-1}$ coincide just *inside* of the former's leftmost zero, which means that a small, negative $a_1$ will push $J$'s minimum eigenvalue left, in accord with your conjecture.

This can be extended to $L$ having a nonzero element in its diagonal, which will push its minimum eigenvalue either left or right but no more than $J$'s, at least at first order.

Since your result is (or appears to be) true both for big and small $b$, I should expect it to be true everywhere, or to have some pretty interesting mathematics in the middle.