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 $(x-1)\mathrm e^{-ay}$.