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. In fact we can check that $F(C(D))\subset W^{1,\infty}(D)$, which will be useful later. 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.
Proving $W^{1,\infty}$ follows from the range of $F$, and the fact that $g$ is itself in $W^{1,\infty}$ for $k=0$.
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.