If the experiment was set up as shown in the image below, where the two slits are at different distances from the screen,
would the fringe spacing still be y = ($\lambda)*L/d$ where $\lambda$ is the wavelength of light, $d$ is the slit spacing (along the y-axis) and $L$ is the distance from the furthest slit to the screen? If not, what would the fringe spacing equation now be?
This problem is equivalent to set your screen slightly rotated an angle $\alpha$ (see the figure below).
If your spacing is $a$, then your new spacing with the rotated screen would be $a' = a \cos \alpha$.
Edit: You can take the approximation $\cos \alpha \approx 1 - \frac{\alpha^2}{2}$, which is valid for $\alpha \ll 1$. You can see that there isn't any first-order term, so this means that if you have small enough angles the $\alpha^2$ term is negligible. This proves that, at first order approximation, the spacing would be the same, but this is only true if $\alpha$ is small enough. This is the situation when the two slits are almost aligned. This justifies S. McGrew's answer.
Offsetting one of the slits in the direction of L by a small amount just shifts the fringe positions, without affecting fringe spacing.
