You need $A$ to be finite-dimensional to define the trace.
Given that, it's a classic exercise to show that if the trace form is nondegenerate then $A$ is semisimple; this is done in this math.SE answer using the fact that the Jacobson radical must be contained in the radical of the trace form, so if the trace form is nondegenerate then the Jacobson radical vanishes. So $A$ is Artinian + semiprimitive which is equivalent to semisimple.
Next we argue that the nondegeneracy of the trace form passes to extensions of scalars $A \otimes_k L$. Maybe the most concrete way to see this is to note that nondegeneracy is equivalent to the determinant of a Gram matrix $\det \text{tr}(e_i e_j)$ being nonzero where $\{ e_i \}$ is a basis, and this condition is clearly preserved (and reflected) by extension of scalars. So now it follows that if the trace form is nondegenerate then $A \otimes_k L$ is semisimple for every field extension $k \to L$; this says that $A$ is geometrically semismple.
To prove the converse we note that the argument above showing that nondegeneracy of the trace form passes to extension of scalars also shows that nondegeneracy of the trace form can be checked by checking it on any extension of scalars; in particular it can be checked for the extension of scalars $A \otimes_k \bar{k}$ to the algebraic closure. By Artin-Wedderburn we have
$$A \otimes_k \bar{k} \cong \prod_i M_{n_i}(\bar{k})$$
and we can check by hand that the trace form is nondegenerate for any such algebra. Alternatively, $M_n(k)$ is simple so has no nontrivial two-sided ideals, but the radical of the trace form is a two-sided ideal and the trace form is nonzero, so the radical must be zero. And the direct product of two algebras whose trace forms are nondegenerate again has nondegenerate trace form.
It's also worth knowing that the geometrically semisimple $k$-algebras can be classified exactly:
Theorem: A finite-dimensional $k$-algebra $A$ is geometrically semisimple iff it is a finite product $\prod_i M_{n_i}(D_i)$ of matrix algebras over finite-dimensional division algebras $D_i$ whose centers are finite separable extensions of $k$.
For a proof see this blog post. In particular if $A$ is commutative then it's geometrically semisimple iff it's a finite product of finite separable extensions of $k$, which says exactly that $A$ is an étale algebra over $k$, so we recover as a special case that a finite field extension $L/k$ is separable iff it's geometrically semisimple iff $L \otimes_k \bar{k}$ is isomorphic to a finite product of copies of $\bar{k}$ iff $L \otimes_k \bar{k}$ has no nontrivial nilpotents iff the trace form on $L$ is nondegenerate.
