The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:\mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $\mathcal O_L/(p) \cong \mathbf F_p^r$ but if $\mathcal O_L/(p) \cong \mathbf F_p[X]/(f)$ for monic $f$ in $\mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $\mathcal O_L$ is not $\mathbf Z[\alpha]$ for some $\alpha$ ($L$ is not monogenic).

Dedekind found the first example of this: $L = \mathbf Q(\theta)$ where $\theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $\mathcal O_L/(2) \cong \mathbf F_2^3$ and $\mathcal O_L/(2) \not\cong \mathbf F_2[X]/(f)$ for any $f$ in $\mathbf F_2[X]$.

The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:\mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $\mathcal O_L/(p) \cong \mathbf F_p^r$ but if $\mathcal O_L/(p) \cong \mathbf F_p[X]/(f)$ for monic $f$ in $\mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $\mathcal O_L$ is not $\mathbf Z[\alpha]$ for some $\alpha$ ($L$ is not monogenic).

Dedekind found the first example of this: $L = \mathbf Q(\theta)$ where $\theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $\mathcal O_L/(2) \cong \mathbf F_2^3$ and $\mathcal O_L/(2) \not\cong \mathbf F_2[X]/(f)$ for any $f$ in $\mathbf F_2[X]$. Also $\mathcal O_L \not= \mathbf Z[\alpha]$ for some $\alpha$ in $\mathcal O_L$.

The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:\mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $\mathcal O_L/(p) \cong \mathbf F_p^r$ but if $\mathcal O_L/(p) \cong \mathbf F_p[X]/(f)$ for monic $f$ in $\mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $\mathcal O_L$ is not $\mathbf Z[\alpha]$ for some $\alpha$ ($L$ is not monogenic).

Dedekind found the first example of this: $L = \mathbf Q(\alpha)$ where $\alpha$ is a root of $X^3-X^2-2X-8$. This a cubic field in which $2$ splits completely. Thus $\mathcal O_L/(2) \cong \mathbf F_2^3$ and $\mathcal O_L/(2) \not\cong \mathbf F_2[X]/(f)$ for any $f$ in $\mathbf F_2[X]$.

The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:\mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $\mathcal O_L/(p) \cong \mathbf F_p^r$ but if $\mathcal O_L/(p) \cong \mathbf F_p[X]/(f)$ for monic $f$ in $\mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $\mathcal O_L$ is not $\mathbf Z[\alpha]$ for some $\alpha$ ($L$ is not monogenic).

Dedekind found the first example of this: $L = \mathbf Q(\theta)$ where $\theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $\mathcal O_L/(2) \cong \mathbf F_2^3$ and $\mathcal O_L/(2) \not\cong \mathbf F_2[X]/(f)$ for any $f$ in $\mathbf F_2[X]$.