How did Einstein integrate $\frac{\partial \tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial \tau}{\partial t}=0$?

From the definitions given: $$\partial_{x'}\tau = A, \quad \partial_t \tau = D$$ Also: $$\partial_{y}\tau = B = 0, \quad \partial_z \tau = C = 0$$ From the differential equation we get: $$A + \frac{v}{c^2 - v^2}D = 0 \\ \implies A = - \frac{v}{c^2 - v^2}D$$ Now using the definition given for $\tau$: $$\tau = - \frac{v}{c^2 - v^2}Dx' + Dt = D\bigg(t - \frac{v}{c^2 - v^2}x'\bigg)$$ So if you define $D=a$ you get the final expression for $\tau$. As you have noticed, requiring $\tau$ to be affine doesn't change the results at all - your $\tau$ would be: $$ \tau = a\bigg(t - \frac{v}{c^2 - v^2}x'\bigg) + E$$ The reason $a=a(v)$ is because $D$, who is actually $a$ in disguise, does not depend on $x', y, z, t$ but is assumed to depend on $v$. Otherwise, the relation wouldn't be linear. Without further information, $a$ is a function of potentially anything except those four variables.

There is really no need to assume that $\tau$ is linear or affine to derive the general form of $\tau$. Write the equation as $$ \frac{\partial\tau}{\partial x'}+k\frac{\partial\tau}{\partial t} = 0 , $$ where $k={v}/({c^2-v^2})$. Change coordinates from $(x',t)$ to $(\xi,u)$ by $$ \begin{cases}\xi= x'\\ u=t-k x' \end{cases} \qquad \text{or equivalently,}\qquad \begin{cases}x'=\xi\\ t=u+k\xi. \end{cases} $$ This gives $$ \frac{\partial\tau}{\partial\xi} = \frac{\partial\tau}{\partial x'} \frac{\partial x'}{\partial\xi}+ \frac{\partial\tau}{\partial t}\frac{\partial t}{\partial\xi} = \frac{\partial\tau}{\partial x'}+k\frac{\partial\tau}{\partial t} = 0 , $$ meaning that in the new coordinates, $\tau$ is a function of only $u$. Hence $$ \tau=a(u)=a(t-k x'), $$ for some function $a$. At this point, you can use the assumption that $\tau$ is linear (or affine) to deduce that $a$ is linear (or affine).