This proof might not count as direct. It is for finite extensions.

Lemma: $L/K$, a finite extension, is Galois if and only if $L \otimes_K L$ is a product of copies of $L$.

Proof: If $L\otimes_K L$ is a product of copies of $L$, and $x$ is fixed by every automorphism, then $s \otimes 1- 1 \otimes s$ is zero in $L \otimes_K L$, so $s \in K$ (via explicit description of tensor products of vector spaces.)

If there are a lot of automorphisms, then each automorphism gives a different surjective map $L \otimes_K L \to L$, so we get a surjective map from $L \otimes_K L $ to the product of $[L:K]$ copies of $L$, which must be an isomorphism by dimension-counting.

Then $L \otimes_M L$ is a quotient of $L \otimes_K L$, so is a product of finitely many copies of $L$.

I believe one can extend this to all algebraic extensions via a slightly more complicated argument. But that might just be pointless, as one could argue that my condition is just a clever way of saying "normal and separable" in different language.

This proof might not count as direct. It is for finite extensions.

Lemma: $L/K$, a finite extension, is Galois if and only if $L \otimes_K L$ is a product of copies of $L$.

Proof: If $L\otimes_K L$ is a product of copies of $L$, and $s$ is fixed by every automorphism, then $s \otimes 1- 1 \otimes s$ is zero in $L \otimes_K L$, so $s \in K$ (via explicit description of tensor products of vector spaces.)

If there are a lot of automorphisms, then each automorphism gives a different surjective map $L \otimes_K L \to L$, so we get a surjective map from $L \otimes_K L $ to the product of $[L:K]$ copies of $L$, which must be an isomorphism by dimension-counting.

Then $L \otimes_M L$ is a quotient of $L \otimes_K L$, so is a product of finitely many copies of $L$.

I believe one can extend this to all algebraic extensions via a slightly more complicated argument. But that might just be pointless, as one could argue that my condition is just a clever way of saying "normal and separable" in different language.