How can I derive an approximate equation for the forward voltage drop of a particular component?
You might be able to skip curve fitting all together by examining the vendor's SPICE model for this device:

**********************************
* Copyright:                     *
*   Thomatronik GmbH, Germany    *
*   info@thomatronik.de          *
**********************************
*   SPICE3
.subckt ssa34 1 2
ddio 1 2 legd
dgr 1 2 grd
.model legd d is = 1.73647E-006 n = 1.01682 rs = 0.0395584
+ eg = 0.4 xti = 0.200031 tnom = 27
+ cjo = 8.10972E-010 vj = 0.700021 m = 0.588084 fc = 0.5
+ tt = 1.4427E-009 bv = 44 ibv = 1 af = 1 kf = 0
.model grd d is = 6.76885E-008 n = 0.972506 rs = 0.0698092
+ eg = 1 xti = 0.2 tnom = 27
.ends


From the is and n terms of the legd model, you can get the dc behavior as

\$I(V) = I_s(\exp(\frac{V}{nV_t})-1)\$

The rs terms shows about 40 mOhms of series resistance, which might explain why your pure logarithmic fit is not perfect at high currents.

You can invert the diode equation and add the series resistance term to get the solution you want:

\$V(I) = nV_t\ln(I/I_s + 1) + IR_s\$

Hopefully this will match (nearly) the curve you have been trying to fit.

Also notice that they had to model the device as two parallel diodes (I am guessing one is a parasitic path and only has a small effect on the overall behavior), so it may not be possible to get a perfect fit to a single exponential curve over a wide current range for this device, even with the series resistance term.

From some spot calculations, the dgr diode contributes about 5%-6% of current at reasonable operating points (0.1 - 1 A). You can probably ignore it, because this level of error will probably be lower than your error for not knowing the operating temperature of the diode.