If you're asking about the table on page 2 (on which figure 2 ain't located), that's beacuse the datasheet covers multiple physical devices, which are actually called that way, i.e have that suffix on their part number. Only the one without any [voltage] suffix is the "officially" adjustable one. But as The Photon explained the difference is somewhat academic, because the adjustable one simply sets its output [typically] 1.25V below its ADJ

Also the 1117 is made in similar versions by other companies. It sometimes helps to read the datasheets of bigger manufacturers, which will [usually] put more effort in an understandable datasheet. For this device I like ON's datasheet in particular... because it has more data than the others (you'll see that even more clearly when you get to the output cap), and also explains things clearly from the outset:
This series contains nine fixed output voltages of 1.5 V,
1.8 V, 1.9 V, 2.0 V, 2.5 V, 2.85 V, 3.3 V, 5.0 V, and 12 V that have no
minimum load requirement to maintain regulation. Also included is an
adjustable output version that can be programmed from 1.25 V to
18.8 V with two external resistors.
Note that the adjustable [AMS]1117 does have minimum load requirement. From the AMS datasheet (page 4), that's 10mA.
Also the first two sentences of the following statement in the AMS1117 datasheet (p. 4) are rather misleading because they don't get explained further therein:
The AMS1117 series develops a 1.25V reference voltage between the output and the adjust terminal. Placing a resistor between these
two terminals causes a constant current to flow through R1 and
down through R2 to set the overall output voltage. This current is
normally the specified minimum load current of 10mA. Because
IADJ is very small and constant it represents a small error and it
can usually be ignored.
Compare with what ON's datasheet says on this:
The typical application circuits for the fixed and adjustable output regulators are shown in Figures 23 and 24. The adjustable devices are floating voltage regulators. They develop and maintain the nominal 1.25 V reference voltage between the output and adjust pins. The reference voltage is
programmed to a constant current source by resistor R1, and
this current flows through R2 to ground to set the output
voltage. The programmed current level is usually selected to
be greater than the specified 5.0 mA minimum that is
required for regulation. Since the adjust pin current, Iadj, is
significantly lower and constant with respect to the
programmed load current, it generates a small output
voltage error that can usually be ignored. For the fixed
output devices R1 and R2 are included within the device and
the ground current Ignd, ranges from 3.0 mA to 5.0 mA
depending upon the output voltage.

Also there is nothing like ON's Fig 23 in the AMS datasheet, even though that's how you'd use the AMS1117-1.5 though AMS1117-5.0 [fixed output] devices, a lacunae that understandably adds to newbie confusion.
Also NXP's diagram of the internals of theirs, which is generic/general helps fix this idea:

And if you're really curious, you can "cheat" and adjust the output of the fixed
ones too as shown in LT's datasheet:

This is going to hurt, but I am guessing you forgot to pull the EN pin high. (or you did pull it high but there is a broken connection somewhere)
Here's the chip block diagram from the datasheet:

With the EN pin low, the boost converter is disabled but the LDO portion not affected (at least, if the block diagram is honest). Then DC conduction through the inductor and the switching diode (between the SW and BSTOUT pins) will produce a voltage on BSTOUT about 0.5 to 0.7 V below the input voltage (like you saw). This would normally put the LDO into dropout, but when you get this voltage sufficiently above 5 V, the LDO will work normally. Again, this is exactly the behavior you observed.
If you just left the EN pin floating, there is nothing in the datasheet that says which way the pin will default. So you will need to actively pull it high to enable the chip.
Best Answer
You could use an LM7705 and three tiny ceramic caps to generate a -0.232 volt negative rail and then you'd be able to get down to about +1V.
\$V_{OUT} = V_{REF} (1 + \frac{R_2}{R_1}) - 0.232 +I_{ADG}R_2\$
So for R2 = 0, Vout ~= 1.02V
simulate this circuit – Schematic created using CircuitLab
Where \$R_2 (\Omega) \approx 96.2(V_{OUT} -1.018)\$
Or use another type of regulator (what I would suggest if the neg rail isn't otherwise useful).