The method of characteristics yields the equations
$ \begin{eqnarray} \dot x &=& a(x-1)\;,\\ \dot y &=& 1\;,\\ \dot z &=& 0\;, \end{eqnarray} $
with solutions
$ \begin{eqnarray} x - 1 &=& c_1\mathrm e^{at}\;,\\ y &=& t + c_2\;,\\ z &=& c_3\;. \end{eqnarray} $
So the function value $z=f(x,y)$ is constant along these curves. We can combine the solutions for $x(t)$ and $y(t)$ into an equation
$(x-1)\mathrm e^{-ay}=c_4\;.$
This tells us that $f$ only depends on $x$ and $y$ through $(x-1)\mathrm e^{-at}$, which is Didier's solution in the comments.
Yes, you can also solve this by separation of variables. The ansatz $f(x,y)=X(x)Y(y)$ leads to
\frac{X'}{X}a(x-1)=-\frac{Y'}{Y}\;.
Setting both sides equal to a constant $c$ gives you two ordinary differential equations, with solutions
$ \begin{eqnarray} X(x)&=&c_1(x-1)^{\frac{c}{a}}\;,\\ Y(y)&=&c_2\mathrm e^{-cy}\;. \end{eqnarray} $
That gives you solutions
$ \begin{eqnarray} f(x,y)&=&c_3(x-1)^{\frac{c}{a}}\mathrm e^{-cy}\\ &=&c_3\left((x-1)\mathrm e^{-ay}\right)^{\frac{c}{a}}\;. \end{eqnarray} $
Since you can choose $c$ freely, you can combine these into arbitrary functions of $(x-1)\mathrm e^{-ay}$.