I've tried the following :
$$\Bbb Z_2[X]/\langle X^2+X\rangle = \{ a + bX + \langle X^2 +X \rangle \mid a,b \in \Bbb Z_2\}$$
The possibilities are $a = 0,b\in \{0,1\}$ and $a=1, b\in \{0,1\}$. As such, there are 4 elements in this factorisation.
Is my reasoning correct?
Yes you are correct since $0$, $1$, $X$ and $X+1$ are the only polynomials of degree less than $2$ in $\mathbb{Z}_2[X]$.
Let's suppose that there was another element in the ring $\mathbb{Z}_2[X]/{\langle X^2+X\rangle}$ from the ones that you stated above $$f+\langle X^2+X \rangle=X^n+a_{n-1}X^{n-1}+\ldots +a_1X+a_0+\langle X^2+X \rangle$$ where $n\geq 2$, then by the division algorithm $$f=X^n+a_{n-1}X^{n-1}+\ldots +a_1X+a_0=q(X^2+X)+r$$ with the degree of $r$ being less than $2$.
The polynomial $r$ is one of the four elements stated above. The polynomial $f-r$ is in the ideal $\langle X^2+X\rangle$.
We can then conclude that $f+\langle X^2+X \rangle=r+\langle X^2+X \rangle$ and that the ring $\mathbb{Z}_2[X]/{\langle X^2+X\rangle}$ cannot have more than four elements.
The four elements in the ring are all distinct since the difference between two different polynomial of degree less than $2$ is a non-zero polynomial of degree less than $2$. Let $f,g$ be two distinct polynomial of degree less than $2$ and $d=f-g$. Let's suppose that $d$ is in the ideal generated by $X_2+X$, then there would exist a polynomial $q$ such that $d=q(X^2+X)$.
However, $d=0\cdot(X^2+X)+d$ is a valid euclidean division on the polynomial $d$ since the degree of $d$ is less than $2$. Since there should be unique $q,r\in \mathbb{Z}_2[X]$ such that $d=q(X^2+X)+r$ with the degree of $r$ being less than $2$, we can conclude that $d$ is not in the ideal generated by $X^2+X$.
This means that all the polynomials of degree less than two are representatives of the differents cosets of the ideal. The quotient ring has exactly four elements.