To answer a request in comments : let us show that there is existence and uniqueness on $D_0=D\cap\{(x,y):y(x-y)\| f\|_{L^\infty((0,1)^2)}<1-\epsilon\}$, so for example on $D$ when $\|f\|_{L^\infty((0,1)^2)}<4$.
Let us show 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 $$ \left| F(u)(x,y) \right|\leq \|u\|_\infty \|f\|_\infty \left|\int_0^y \int_0^{x-y} d s d \tau \right| = \|u\|_\infty \|f\|_\infty y(x-y) < \|u\|_\infty(1-\epsilon) $$ 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. Proving $W^{1,\infty}$ regularity after that is just calculus.
Regarding $\|f\|_\infty<4$ : $y(x-y)=x^2 t(1-t)$ with $t\in(0,1)$ so $y(x-y)<\frac{x^2}{4}<\frac14$. Therefore $\|Fu\|_\infty <\frac14 \|f\|_\infty<1$, and that's that.