1. Let $I=\int_0^2 f(t)\,dt.$ Since the integral is over a fixed interval, $I$ is a constant.

2. The given differential equation is $f'(x)+f(x)=I.$ So we solve the linear ODE $f'(x)+f(x)=I.$

3. Solve the ODE.

   The complementary solution is $f_c(x)=Ce^{-x}.$

   Since the RHS is a constant, a particular solution is also a constant, say $f_p(x)=I$.

   Hence the general solution is $f(x)=I+Ce^{-x}.$

4. Use the condition $f(0)=e^{-2}$: $f(0)=I+C=e^{-2}. \qquad (1)$

5. Now use the definition of $I$: $I=\int_0^2 f(t)\,dt=\int_0^2 \left(I+Ce^{-t}\right)dt.$

   Therefore, $I=2I+C\int_0^2 e^{-t}dt=2I+C(1-e^{-2}).$

   So, $I=2I+C(1-e^{-2})$ $0=I+C(1-e^{-2})$ $I=-C(1-e^{-2}). \qquad (2)$

6. Substitute (2) into (1): $-C(1-e^{-2})+C=e^{-2}$ $C\big(1-(1-e^{-2})\big)=e^{-2}$ $Ce^{-2}=e^{-2}$ $C=1.$

   Then from (2), $I=-(1-e^{-2})=e^{-2}-1.$

7. Hence $f(x)=I+e^{-x}=e^{-2}-1+e^{-x}.$

   Now compute $f(2)$: $f(2)=e^{-2}-1+e^{-2}=2e^{-2}-1.$

8. Therefore, $2f(0)-f(2)=2e^{-2}-(2e^{-2}-1)=1.$

So the required integer is $\boxed{1}.$