When a floating-gate transistor has no negative charge in the gate (i.e., it is erased), it conducts when the world-line voltage is applied to its gate terminal, pulling the bit-line to ground. Doesn't this mean that an erased cell represents a logical '0' rather than a '1'?
No. Which voltage you interpret as 0 or 1 is completely up for interpretation. Humans tend to thing "higher voltage == 1", but there's no technical reason for that. You can map low voltage to 1 as much as you like, or you can map much larger blocks of cell states to blocks of bits:
In fact, for flash memory of any significant size, you'll find that error-correcting codes are used to allow for smaller cells whilst staying reliable. With that, there's code bits that get mapped to data bits in a block code – and how that code looks like defines how "block of flash memory cell states" map to "block of data bits"; this doesn't only involve arbitrary inversions (and non-inversions), but extends to non-systematic codes: It's pretty possible that if you change a single bit on e.g. an SSD, there's no single cell that always changes, regardless of the other bits in the block.
I'd recommend mentally separating "charge of a capacitor" from "bit": The bit's value is an interpretation of that charge, anyways.
And that's a pretty important property of flash memory: there's inherently a comparison of voltages / currents to a reference voltage/current, which is used to decide in which state the memory cell is when reading.
