As to my understanding in order to calculate the behavior of the extrema of a function, a good approach is to examine the behavior of the first derivative of the given function. So by calculating the derivative I got: $$ f'(x)=(e^x(x^2+a))'=e^x(x^2+2x+a)$$
now in order to examine the behavior of $f(x)$ I examine where $f'(x)$ is positive or negative. And at the points where $f'(x)=0$ there will be an extremum.
So because $$ e^x >0 \space \forall x\in \mathbb R $$ whether $f'(x)$ is negative or positive is defined by the polynom $x^2 +2x+a$
Then by using the quadratic formula I found that for the discriminant $\Delta$ the following are true.
For $a<1$ it's $\Delta > 0 $ that means that for $a \in(-\infty,1)$, $f(x)$ has two extrema because the polynom has two roots and therefore the monotony of $f(x)$ changes twice.
And for $ a > 1$ it's $\Delta<0 $, the polynom has no real roots and therefore for $a\in(1,+\infty)$, $f(x)$ shows no change in its monotony and as a result it has $0$ extrema.
For the special case of $\Delta = 0$ the polynom is as well everywhere positive and therefore once again $f(x)$ has not extrema.
So there is no value for $a$ so that $f(x)$ will have only one extremum.
The thing is that just examining that way doesn't seem that "mathematically" correct to me and I wonder if there is any better way to examine and actually prove that there is no $a$ so that $f(x)$ has only one extremum.
You've discussed correctly the cases for $a\ne 1$. The case for $a=1$ is not clearly examined. $\Delta=0$ means that the two solutions for the derivative are equal, both at $x=-1$. What you need to do is to show that this is not an extremum, even if the derivative is $0$, but an inflection point. You will need to take the second derivative, and check that it changes sign at $x=-1$. In a way, this problem is equivalent to showing that $x^3$ has no extremum, even if the derivative at $0$ is $0$.