If $A$ is a $k$-algebra, recall that $\Omega_{A/k}$ is $I/I^2$ where $I$ is the kernel of the multiplication map $A \otimes_k A \to A$. If $I$ is a finitely generated ideal of $A \otimes_k A$, and in particular if $A \otimes_k A$ is Noetherian (e.g. if $A$ is essentially of finite type), then $\Omega_{A/k} = 0$ iff $I = I^2$ iff (by Nakayama) $I$ is generated by an idempotent.

This condition implies that $A$ is a projective $A \otimes_k A$-module, so $A$ is separable over $k$, and hence must be a finite product of finite separable extensions of $k$ by the classification of separable algebras. When $k = \mathbb{F}_p$ all such algebras clearly have surjective Frobenius.

I'm not sure what to do about the non-Noetherian case.

If $A$ is a $k$-algebra, recall that $\Omega_{A/k}$ is $I/I^2$ where $I$ is the kernel of the multiplication map $A \otimes_k A \to A$. If $I$ is a finitely generated ideal of $A \otimes_k A$, and in particular if $A \otimes_k A$ is Noetherian (e.g. if $A$ is essentially of finite type), then the following conditions are equivalent:

1. $\Omega_{A/k} = 0$,
2. $I = I^2$,
3. $I$ is generated by an idempotent,
4. $A$ is a projective $A \otimes_k A$-module.

This means that $A$ is separable over $k$, and hence must be a finite product of finite separable extensions of $k$ by the classification of separable algebras. When $k = \mathbb{F}_p$ all such algebras clearly have surjective Frobenius.

I'm not sure what to do about the non-Noetherian case.

If $A$ is a $k$-algebra, recall that $\Omega_{A/k}$ is $I/I^2$ where $I$ is the kernel of the multiplication map $A \otimes_k A \to A$. If $I$ is a finitely generated ideal of $A \otimes_k A$, and in particular if $A \otimes_k A$ is Noetherian (e.g. if $A$ is essentially of finite type), then the following conditions are equivalent:

1. $\Omega_{A/k} = 0$,
2. $I = I^2$,
3. $I$ is generated by an idempotent,
4. $A$ is a projective $A \otimes_k A$-module.

This is exactly the condition that $A$ is separable over $k$, and since $A$ is commutative this means it must be a finite product of finite separable extensions of $k$ by the classification of separable algebras. When $k = \mathbb{F}_p$ all such algebras clearly have surjective Frobenius.

I'm not sure what to do about the non-Noetherian case.

If $A$ is a $k$-algebra, recall that $\Omega_{A/k}$ is $I/I^2$ where $I$ is the kernel of the multiplication map $A \otimes_k A \to A$. If $I$ is a finitely generated ideal of $A \otimes_k A$, and in particular if $A \otimes_k A$ is Noetherian (e.g. if $A$ is essentially of finite type), then $\Omega_{A/k} = 0$ iff $I = I^2$ iff (by Nakayama) $I$ is generated by an idempotent.

This condition implies that $A$ is a projective $A \otimes_k A$-module, so $A$ is separable over $k$, and hence must be a finite product of finite separable extensions of $k$ by the classification of separable algebras. When $k = \mathbb{F}_p$ all such algebras clearly have surjective Frobenius.

I'm not sure what to do about the non-Noetherian case.