To answer a request in comments : let us show that there is existence and uniqueness on $D_0=D\cap\{(x,y):y(x-y)\| f\|_{L^\infty((0,1)^2)}<1-\epsilon\}$, or on $D$ when $\|f\|_{L^\infty((0,1)^2)}<4$.

Let us show existence and uniqueness among continuous functions on $D_0$ equipped with the max norm. 
$$
F:u\to \int_0^y \int_0^{x-y} u(x-y+\tau, s+\tau) f(s+\tau, \tau) d s  d \tau
$$
which acts on $C(D_0)$ into itself.  We compute 
$$
\sup_{u\neq0}\frac{\|Fu\|_\infty}{\|u\|_\infty} \leq \|f\|_\infty \left|\int_0^y \int_0^{x-y}  d s  d \tau \right| = \|f\|_\infty y(x-y).
 < (1-\epsilon) $$
thus, for example, the sequence 
$$
u_{n+1} = F(u_n) + g
$$
with $g=-\int_0^yf(x-y+\tau,\tau)d\tau$ converges to a solution of $u=Fu+g$, since $F$ is a contraction. Explicitly, the solution is
$$
u=\sum_{k=0}^\infty F^k g
$$
where $F^k$ means $F$ composed with $F$ ,$k$ times.


Regarding $\|f\|_\infty<4$ : since $y(x-y) \leq\frac{x^2}{4}<\frac14$,  $$\sup_{u\neq0}\frac{\|Fu\|_\infty}{\|u\|_\infty} \leq\frac14  \|f\|_\infty<1,$$ and that's that. 

____

(Corrected thanks to a remark from Giorgio Metafune) It isn't immediate that $u$ and /or $g\in W^{1,\infty}$ for general $f$.