the elliptic curves with j-invariant zero Let $B\in K^\ast$, where $K$ is a number field. Let $y^2=X^3+B$ be the Weierstrass equation for an elliptic curve $E_B$ over $K$. 
Note that the $j$-invariant of $E$ is zero.
When is $E_B$ isomorphic to $E_{B^\prime}$ over $K$? (Here $B^\prime \in K^\ast$.) Does this happen if and only if $B=B^\prime$? Or does it also happen if $B^\prime = u B$, where $u$ is a unit in $O_K^\ast$?
How do I determine the semi-stable reduction of $E_B$? That is, we know that $E_B$ has potential good reduction. How do I determine $L/K$ such that the elliptic curve $E_{B}\otimes_K L$ has good reduction over $O_L$? 
If $B$ is a unit, then $E_B$ has good reduction. So we may and do assume $B$ is not a unit, i.e., $B$ is a prime.
I have a feeling that we must choose $L$ such that its ramification over the primes dividing $B$ is of the "right" type. But how do I see what the necessary type is?
 A: Your first question is answered in Silverman AEC I Prop X.5.4.iii (I have the old edition, don't know if this matters).
Namely, the elliptic curves corresponding to $B$ and $B'$ are isomorphic if and only if $B$ and $B'$ differ by a 6th power of an element of $K$. In fact, more is true, namely, the set of isomorphism classes of elliptic curves with $j$-invariant $0$ is a torsor for
$$
\frac{K^\times}{{(K^\times)}^6}
$$ 
under the action $d*E_a= E_{ad}$ where $E_c$ denotes the elliptic curve with Weierstrass equation $y^2 = x^3 + c$.
This explicit description also helps tackle your second question. Namely, if we take a field extension that makes $A$ a 6th power (i.e. we adjoin a 6th root of $A$), then there's a model with everywhere good reduction, at least if we ignore primes above 2 and 3. Dealing with those is going to be a pain - note that your model has bad reduction at $2$ and $3$ even if $A$ is a unit, for instance -  but you can solve it by chasing denominators in the formula for the discriminant. I'm not sure if you're going to wind up needing to adjoin a root of $2$ or of $3$ or not.
