Calculation of elastic constants - when do we need to account for Poisson?

The difference can NOT be ignored, even for an isotropic material (and I would guess your organic crystals are not isotropic, which makes things more complicated).

Assuming your pictures are of an isotropic material, if you apply equal and opposite loads normal to the surface and with no other restraints, you will change the shape of the material as in picture C. You can measure the changes in length and width, and calculate Young's modulus and Poisson's ratio directly from the two measurements. (There will also be a change in thickness, but for an isotropic material that doesn't give any new information about the material properties).

On the other hand if you want the deformed shape to be like picture A, you also need to apply loads to all four sides of the material, not just the two loads that you show in the picture. Again, you can calculate $E$ and $nu$ directly from the two applied loads.

Pictures B and D are not physically realistic, because if the loads as shown would act as a couple and cause rigid body rotation of the material. If that is prevented in some way, the restraints will be applying additional loads which create and equal and opposite couple.

Of course in a computer model you don't have any practical problems imposing any boundary conditions you like, but if you consider all the forces on the structure caused by the boundary conditions, you have enough information to find all the material properties.

For more information about the simplest case, look up "plane stress" and "plane strain" in any textbook on strength of materials (for a simple introduction) or continuum mechanics (for a more complete mathematical treatment).

Note that to measure material properties, the method of "applying forces and measuring the deformation" is often not used for practical reasons. A better method is often to measure the mode shapes and frequencies of several different vibration modes of a test piece. The only physical properties you need to measure are then the undeformed shape of the test specimen, and its density, both of which are easy to measure accurately.

The elasticity parameters reduces to modulus of elasticity $E$ and Poisson coeficient $nu$ only for isotropic materials.

In this case for example:

$sigma_{xx}=(lambda + 2mu)epsilon_{xx} + lambdaepsilon_{yy}+lambda*epsilon_{zz}$
$sigma_{xy}= 2muepsilon_{xy}$

where $lambda$ and $mu$ are the Lamé constants of the material, ($E$ and $nu$ can be expressed as a function of them).

So, measuring the principal strains with a strain gage, and the stresses with a load cell, it is enough two tests, one of them uniaxial, to determine the parameters.

But for a totally anisotropic material, there are 21 elastic parameters, and many different tests are certainly necessary .