1. Let $n=[e^x].$ Since $e^x>0$ for all real $x$, we have $n\in \{0,1,2,\dots\}$.

2. The given equation is $[e^x]^2+[e^x+1]-3=0.$ Using $n=[e^x]$, this becomes $n^2+[e^x+1]-3=0.$

3. Now use the property: $[y+1]=[y]+1$ for every real $y$. Hence, $[e^x+1]=[e^x]+1=n+1.$

4. So the equation reduces to $n^2+(n+1)-3=0$ $n^2+n-2=0$ $(n+2)(n-1)=0.$

5. Therefore, $n=-2 \quad \text{or} \quad n=1.$ But $n=[e^x]\ge 0$, so only $n=1$ is possible.

6. Thus, $[e^x]=1.$ This means $1\le e^x<2.$

7. Taking natural logarithm throughout: $0\le x<\ln 2.$

8. Therefore the set of all such $x$ is $[0,\ln 2).$

9. Compare with the options:

   • A: $\left[0,\frac1e\right)$ — incorrect
   • B: $[\ln 2,\ln 3)$ — incorrect
   • C: $[1,e)$ — incorrect
   • D: $[0,\ln 2)$ — correct

Hence the correct option is D.