It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous).
So we know a priori that Tau is of the form: Tau(x',y,z,t) = Ax' + By + Cz + Dt + E,
and Tau(0,0,0,t) = Dt+E, where D is del(Tau)/del(t).  Likewise all the coefficients are simply the respective partial derivatives: A=del(Tau)/del(x'), B=del(Tau)/del(y), C=del(Tau)/del(z), and E is an additive constant.   
So taking Einstein's 1/2[Tau_0 + Tau_2] = Tau_1 formula, plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation:
del(Tau)/del(x') + (v/(c^2-v^2))*del(Tau)/del(t) = 0.  Note that during this algebraic process x' simply cancels out anyway.