Let $E$ be an elliptic curve over a (perfect) field $K$ and let $G_K = \mathrm{Gal}(\overline K/K)$. As explained in Chapter X.2 of Silverman's Arithmetic of Elliptic Curves, the twists of $E$ are in bijective correspondence with elements of $H^1(G_K, \mathrm{Aut}(E))$.
In general, $\mathrm{End}(E) = \mathbb Z$, so $\mathrm{Aut}(E) = \{\pm1\}$. Since $G_K$ acts trivially on $\{\pm1\}$, in this case,
$$H^1(G_K, \mathrm{Aut}(E)) = \mathrm{Hom}(G_K, \{\pm1\})$$
i.e. the twists of $E$ are in bijective correspondence with quadratic Galois characters $G_K\to\{\pm 1\}$. Alternatively, by Kummer theory, $H^1(G_K,\{\pm 1\})\cong K^\times/K^{\times 2}$, which explains why we usually talk about twisting by an element $d\in K^\times$. The character $\chi\colon G_K\to\{\pm 1\}$ corresponding to $d$ is exactly the lift of the canonical character $\mathrm{Gal}(K(\sqrt d)/K)\to \{\pm 1\}$.
If $\mathrm{Aut}(E)$ is strictly larger than $\{\pm1\}$, there will be extra types of twists. When $K$ is a number field, since $\mathrm{End}(E)$ is at worst an order in an imaginary quadratic field, the only way in which $E$ can gain extra automorphisms is if $\mathrm{End}(E) = \mathbb Z[i]$ or $\mathbb Z[\zeta_3]$. This happens exactly for the elliptic curves of $j$-invariant $1728$ and $0$.
In the case of a $j$-invariant $1728$ curve (e.g. take $E\colon y^2 = x^3 - x$), we have $\mathrm{End}(E) = \mathbb Z[i]$, so $\mathrm{Aut}(E) = \{\pm 1, \pm i\} = \mu_4$. Kummer theory gives an isomorphism
$$H^1(G_K, \mu_4)\cong K^\times/K^{\times 4}$$
and the twist corresponding to $d\in K^\times/K^{\times 4}$ is $E_d\colon y^2 = x^3 - dx$. Note that, over $K(\sqrt[4]d)$, the map $(x, y)\mapsto (\sqrt dx,\sqrt[4]{d^3}y)$ identifies $E$ and $E_d$. In particular, if $d$ is a square, then $E_d$ is just a quadratic twist.
If $K\supset \mathbb Q(i)$, then $G_K$ acts trivially on $\mu_4$, and again $H^1(G_K, \mu_4)\cong\mathrm{Hom}(G_K, \mu_4)$ so that we get a bijection between twists and quartic Galois characters $G_K\to\mu_4$. The character corresponding to $d\in K^{\times}$ is exactly the map $\sigma\mapsto \frac{\sigma(\sqrt[4]d)}{\sqrt[4]d}\in\mu_4$.
In the case of $j$-invariant $0$ (e.g. $E\colon y^2 = x^3 + 1$), $\mathrm{End}(E) = \mathbb Z[\zeta_3] = \mathbb Z[\zeta_6]$ and you get sextic twists and twists by sextic characters $G_K\to\mu_6$ (if $K\supset \mathbb Q(\zeta_3)$) in the same way. I don't know why the Wikipedia article focuses on just the cubic twists in this case.
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
No, the $j$-invariant only classifies elliptic curves over an algebraically closed field. Curves with the same $j$-invariant may not be isomorphic over $k$ as they could be twists. That is, a curve and its twist might not be isomorphic over $k$, but there is an isomorphism defined over some extension of $k$.
For instance, consider the elliptic curves $E_1 : y^2 = x^3 - 1$ and $E_2: y^2 = x^3 + 1$. One can show that these elliptic curves are not isomorphic over $\mathbb{Q}$. However, there is an easy isomorphism over $\mathbb{Q}(i)$: \begin{align*} E_1 &\to E_2\\ (x,y) &\mapsto (-x,iy) \, . \end{align*} (Indeed, I cooked up this example by forming the quadratic twist of $E_1$ by $-1$.)