A cheap concave solution on $$\mathbb{R}^2$$ is $$f_0(x,y):=3 \min\Big(\frac{x+1}{3} ,y\Big)-1\, ,$$
which also verifies $$f_0(x,y)=y$$ for all $$(x,y)$$ in the rectangle $$[-1/4, 5/4]\times [-1/4,1/4]=\big([0,1]\times\{0 \}\big)\;{\bf +}\; [-1/4,1/4]^2 \, ,$$ and $$f_0(x,y)=x$$ in the rectangle $$[-1/4, 5/4]\times [3/4,5/4]=\big([0,1]\times\{1 \}\big)\big)\;{\bf +}\; [-1/4,1/4]^2 \, .$$ This implies that if $$\phi$$ is a symmetric $$C^\infty$$ convolution kernel with support in $$[-1/4,1/4]^2$$, the function $$f:=f_0*\phi$$ is a $$C^\infty$$ concave function on $$\mathbb{R}^2$$ and $$f(x,0)=0$$ and $$f(x,1)=x$$.
(We can take e.g. $$\phi(x,y):=\psi(x)\psi(y)$$ with $$\psi\in C^\infty(\mathbb{R})$$, $$\psi\ge0$$, $$\operatorname{supp}(\psi)\subset[-1/4,1/4]$$, $$\int_\mathbb{R}\psi(t)dt=1$$)