The area enclosed between the curve $y={\log }_e(x+e)$ and the coordinate axes is:

The curve is $y=\ln (x+e)$.Intersection with y-axis: $x=0⟹y=\ln (e)=1$. Point is $(0,1)$.Intersection with x-axis: $y=0⟹\ln (x+e)=0⟹x+e=1⟹x=1-e$. Point is $(1-e,0)$.The area enclosed by the curve and coordinate axes is:
$$A={\int }_{1-e}^0\ln (x+e)dx$$
Let $x+e=t⟹dx=dt$.When $x=1-e,t=1$. When $x=0,t=e$.
$$A={\int }_1^e\ln tdt$$
Using integration by parts:
$$A=[t\ln t-t{]}_1^e$$
$$A=(e\ln e-e)-(1\ln 1-1)$$
$$A=(e-e)-(0-1)=1$$