NaOH being added to Formic Acid (Please catch my error!) I am adding 0.55 mL of 0.1M NaOH to 10 mL of 0.5M Formic Acid.

Here are the values I calculated to use.

Moles H+ : 0.5 * (10mL / 1000) = 0.005 mol

Moles OH- : 0.1 * (.55mL / 1000) = 0.000055 mol

volume : (10mL / 1000) + (0.55mL / 1000) = 0.01055 L

Moles Total : moles H - moles OH = 0.005 - 0.000055 = 0.004945 mol

M Total : moles Total / volume = 0.004945 mol / 0.01055 L = .46872 M

Moles Salt : 0.000055 mol

M Salt : moles salt / volume = 0.000055 mol / 0.01055 L = 0.00521 M

Calculations to find the pH

pH = -log(1.80E-4) = 3.745

3.745 - ( -log(M Salt / M Total) ) = 3.745 - 1.95407 = 1.791

However, this value is too low as the lowest pH of 0.5M Formic Acid is ~2.03.

If the volume of NaOH were to say, 0.5 mL, then the M Salt / M Total value would be less than 1/90, and the pH could then be found by 

-log( sqrt( 1.80E-4 * M Total ) ) = 2.036

...which makes more sense, but doesn't apply to the initial problem.

Any idea of where I'm going wrong?
It looks like you misused the acid dissociation constant, which is apparently the $K_a=1.80×10^{-4}$ figure you have.

Properly, you should do the following:

1)  Let $x$ be the hydrogen ion concentration $[\ce{H^+}]$ from the formic acid that actually disdociated.

2)  Since you ended up with $0.469 \text{M}$ total acid the un-dissociated acid $[\ce{HCOOH}]=0.469-x$.

3)  The formate ion comes from both the salt ($0.00521 \text{M}$) and the dissociated acid ($x$), so $[\ce{HCOO^-}]=0.00521+x$.

4)  Plug these into the equilibrium relation

$\frac{[\ce{H^+}][\ce{HCOO^-}]}{[\ce{HCOOH}]}=K_a$

and solve for $x$.  The equation reduces down to a quadratic equation so you should have an easy formula for $x=[\ce{H^+}]$.

5)  Get the pH from the result of Step 4.

Good luck!