Does electron shielding increase or stay constant moving LEFT to RIGHT across a period?
To answer this question, it's important to define what you mean by shielding. Generally, shielding refers to a reduction in the effective nuclear charge experienced by an electron in a given orbital due to the other electrons on the same atom.

The quantitative degree of shielding for a given electron can be approximated by Slater's rules. According to those rules, electrons within the same group of orbitals (for example 4d) have a coefficient of 0.35 (except for 1s electrons, which have 0.30). So valence electrons do shield each other, just not as much as the lower level electrons shield the valence electrons.

For example, let's consider the elements with increasing numbers of 2p electrons (B, C, N, O, F, Ne). Going from left to right, each addition of a 2p electron reduces the effective nuclear charge experienced by another 2p electron by 0.35. So the amount of shielding is increasing as we move left to right.

The apparent contradiction with the ionization energy comes about because you have not considered the increase in the actual nuclear charge. Each time we add a 2p electron, we also add a proton to the nucleus. So even though 0.35 of that charge is shielded from a valence 2p electron, that electron still experiences an increase of 0.65 effective charge and is therefore bound more tightly than a 2p electron on the element to its left.

As a specific case, let's compare the shielding experienced by a 2p electron in carbon versus neon. The carbon electron configuration is $1s^22s^22p^2$. Using Slater's rules, we can calculate that the shielding coefficient for a 2p electron is 2(0.85) + 3(0.35) = 2.75. So the effective nuclear charge experienced by the 2p electron is 6-2.75=3.25.

For neon, the electron config is $1s^22s^22p^6$, so the shielding coefficient of a 2p electron is 2(0.85)+7(0.35)=4.15. So there's more shielding. But the effective charge is 10-4.15=5.85, so the effective charge is still greater and the energy of the coulombic interacation of the electron with the nuclear charge is greater in magnitude. Hence, the ionization energy is greater. 

EDIT: Original version of this answer did not address the question about anion formation (presumably mostly interested in anion formation by addition to an already partially filled shell). When an anion is formed, the additional electron increases the shielding of the other valence electrons, but there is no change in the actual nuclear charge. Thus, each valence electron experiences a lower effective nuclear charge. Since the size of an orbital is a function of the energy of interaction with the nuclear charge, the atomic size increases.