"An instrumentation amplifier is a precision differential voltage gain device [...]." One of the important words here is "gain". An OpAmp has infinite gain (in theory) and only gets a defined gain by adding circuitry around it. Usually, when using one OpAmp only, at least one of the inputs loses its extremely high input impedance because external resistors are necessary.
If you need two (differential) inputs with both a very high input impedance and a defined gain, you can use the two-OpAmp-InAmp you are talking about or the three-OpAmp-InAmp-configuration your picture shows. There are also readymade IC InAmps by such companies as Linear Technology or Analog Devices.
The three-OpAmp-InAmp circuit in in the picture of your question shows that two OpAmps are used as buffers, where they still have a high impedance at their otherwise unconnected non-inverting input pins ("+"). By feeding their outputs into another OpAmp, the upper non-inverting input ("+") becomes an inverting input ("-") because it is connected to the 3rd OpAmp's inverting ("-") input. The lower non-inverting input ("+") remains non-inverting due to its connection with the 3rd OpAmp.
Common three-OpAmp-InAmps use a slightly different configuration compared to your picture to set the gain with one resistor only (the external gain resistor in the case of completely integrated InAmps). Please refer to the links I've provided for more details.
With the three-OpAmp-InAmp, you get both a very high input impedance at two differential inputs (while you would get only one input with such a high input impedance with a regular OpAmp buffer) and you get a very good rejection of common-mode signals (that is achievable with one OpAmp, too, but at the cost of lowering the input impedance with the resistors you have to use to turn the OpAmp into a difference amplifier).
The two-OpAmp-InAmp circuit needs less parts, but at the cost of a not-so-good common mode rejection ratio (CMRR).
Here is a link to a very good book about InAmps by Analog's Charles Kitchin and Lew Counts where you can find a more in-depth look onto all these issues.
Since R1 = R2, for the 2-opamp version the equation for \$V_{OUT}\$ simplifies to
\$V_{OUT} = \left( Sig_+ - Sig_- \right) \times \left( 2 + \dfrac{2 R2}{RG} \right) \$
and indeed there's no sign of R3 or R4. So I made the calculation again, and I found the following, different equation (I don't include the derivation because too much TeX involved):
\$V_{OUT} = \left( Sig_+ - Sig_- \right) \times \left( 2 + \dfrac{R1 + R3}{RG} \right) \$
which I like better because at least we have a term R3 here. Of course if \$R1 = R2 = R3 = R4\$ both equations are equivalent, but this condition isn't mentioned with the schematic. (I'd appreciate it if somebody can confirm that my equation is indeed correct.)
Madmanguruman noted that the gain is minimum 2 for this configuration, which also shows in the above equations. I'm not sure this is a serious restriction, since instrumentation amplifiers are usually used for much higher gains than 2, especially for strain gauge and other Wheatstone bridge measurements. Gains of 100 to 500 are common.
IMO Madmanguruman's other observation that \$Sig_-\$ passes through two opamps is not correct: the inverting input of the top opamp is kept at \$Sig_+\$, and \$Sig_-\$ only influences the currents through the resistors.
It looks like the 2-opamp version is a good alternative for the classic version in most applications, since, like you said, you save an opamp.
edit
In integrated form you don't gain (no pun intended) much from choosing a two-opamp version. The INA122 costs USD 6.86 while the three-opamp INA129 costs USD 7.35, both Digikey prices.
Best Answer
The simplest InAmps need a gain setting resistor and a reference pin like this: -
Now that doesn't look much like an op-amp to me. However, if it doesn't use a gain setting resistor (or maybe it's internally fixed) and, the reference pin is somehow internally dealt with you can use an op-amp symbol.
However, I don't believe there are any InAmps that can internally deal with the reference pin other than by connecting it to midrail and this requires a 0V connection: -
Normally the 0V connection on R3 would exit the chip on a pin.
You have to consider why this is so; An InAmp takes the difference in voltage levels between the two inputs, amplifies it by the gain factor and then references this amplified difference to (for want of better words) to a "reference pin".
This is what an InAmp does.