You really don't need to use $L$-functions or representations. Let's work over a field $K$ of characteristic not equal to 2 or 3, so for example $K=\mathbb Q$. Then an elliptic curve $E/K$ always has a Weierstrass model
$$ E:y^2=x^3+Ax+B, $$
but the model is not unique. The $j$-invariant
$$ j(E) = 1728\cdot\frac{4A^3}{4A^3+27B^2} $$
classifies $E$ up to $\overline K$ isomorphism. You're interested in $K$-isomorphism. Assuming that $j(E)\ne0$ and $j(E)\ne1728$ (i.e., assume that $AB\ne0$), define a new invariant
$$ \gamma(E) = B/A \bmod{{K^*}^2} \in K^*/{K^*}^2. $$
One can check that $\gamma(E)$ is well-defined modulo squares in $K$. Then
$$ \text{$E\cong E'$ over $K$}
\quad\Longleftrightarrow\quad
\text{$j(E)=j(E')$ and $\gamma(E)=\gamma(E')$.}
$$
If $j(E)=0$, then $A=0$ and there's a similar criterion in terms of $B$ modulo ${K^*}^6$, and if $j(E)=1728$, then $B=0$ and there's a criterion in terms of $A$ modulo ${K^*}^4$.
However, probably the right way to understand this is to use the fact that for a given $E/K$, the collection of $E'/K$ that are $\overline{K}$-isomorphic to $E$ are classified by the cohomology group
$$ H^1\bigl(\operatorname{Gal}(\overline K/K),\operatorname{Aut}(E)\bigr). $$
The three cases correspond to $\operatorname{Aut}(E)$ being $\mu_2$, $\mu_6$, and $\mu_4$, respectively, and one knows (Hilbert Theorem 90) that
$$ H^1\bigl(\operatorname{Gal}(\overline K/K),\mu_n\bigr)\cong K^*/{K^*}^n. $$
This unifies the three cases, and gives a quite general way to describe the $\overline{K}/K$-twists of an algebraic variety.
This is too long for a comment. If it doesn't answer your question, please let me know. I also wrote this in a rush, so please double check my claims.
I don't know if you are focused on characteristic 2 or 3, but I think if $k$ is of characteristic $p>3$ then for elliptic curves $E_i$ for $i=1,2$ over $k$ one has that $(E_1)_{\overline{k}}\cong (E_2)_{\overline{k}}$ iff $(E_1)_{k^\mathrm{sep}}\cong (E_2)_{k^\mathrm{sep}}$. The reason is 'simple' (but uses heavy machinery). Consider the functor
$$\mathrm{Isom}(E_1,E_2):\mathbf{Sch}_{/k}\to\mathbf{Set},\qquad T\mapsto \mathrm{Isom}_{\mathbf{Sch}_{/T}}((E_1)_T,(E_2)_T)$$
If $E_1$ and $E_2$ are isomorphic over $\overline{k}$ then, in particular, it's not hard to see that $\mathrm{Isom}(E_1,E_2)$ is a fpqc torsor for $\mathrm{Aut}(E_1)$. This is a finite group scheme over $\mathrm{Spec}(k)$, and since affine morphisms satisfy fpqc descent (see Tag 0245) we know that the $\mathrm{Aut}(E_1)$-torsor $\mathrm{Isom}(E_1,E_2)$ is representable by some $k$-scheme $I$. But, since $\mathrm{char}(k)>3$ we know that $|\mathrm{Aut}(E_1)|$ is invertible in $k$ (cf. [Silv, Theorem 10.1]) and so we know that $\mathrm{Aut}(E_1)$ is smooth over $\mathrm{Spec}(k)$ (e.g. see [Mil, Corollary 11.31]), and so $I$ must also be smooth over $\mathrm{Spec}(k)$ (e.g. see Tag 02VL). But, then $I(k^\mathrm{sep})\ne\varnothing$ (see [Poon, Proposition 3.5.70]). Thus, $(E_1)_{k^\mathrm{sep}}\cong (E_2)_{k^\mathrm{sep}}$.
So, asuming $\mathrm{char}(k)>3$ one can use usual Galois descent arguments to classify elliptic curves over $k$ (e.g. see Theorem 27 of my blog post here).
For characteristic $p=2,3$, the same thing works for classifying forms of $E$ assuming that $p\nmid |\mathrm{Aut}(E)|$. If $p\mid |\mathrm{Aut}(E)|$ I'm not sure what happens.
References
[Mil] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
[Poon] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..
[Silv] Silverman, J.H., 2009. The arithmetic of elliptic curves (Vol. 106). Springer Science & Business Media.
Best Answer
If your finite field $K$ has characteristic $p\ge5$, then you can always put your elliptic curve in the form $$ E_{A,B} : Y^2 = X^3 + AX + B. $$ Then as long as $j(E_{A,B})\ne0,1728$, there are exactly two $K$-isomorphism classes of elliptic curves with that $j$-invariant, namely $E_{A,B}$ and $E_{c^2A,c^3B}$, where $c$ is a quadratic non-residue in $K$, i.e., $c$ represents the non-trivial element in $K^*/(K^*)^2$. For $j=0,1728$, it's a bit more complicated, you'll get non-isomorphic twists for each element of $K^*/(K^*)^6$, $K^*/(K^*)^4$, respectively. And for $p=2$ and $3$, it's even more complicated. In any case, this is all covered in the theory of twists of elliptic curves, see for example Chapter X, Section 5, of my Arithmetic of Elliptic Curves.