If the cardinality of two sets is the same, then when are the two sets equal? 
Take $A, B$ as finite sets. As graciously explained below, this question fails for infinite sets.
  What are all conditions that effect: $|A| = |B| \implies A = B$ ?
  By inspection, the converse is trivially true. 

For example, if $A \subseteq B$ is also true, then the given that $|A| = |B|$ effects  $A = B.$   
For another way to construe this, I try a proof by contrapositive.
Since $A = B \quad  \equiv \quad [A\subseteq B] \quad \wedge \quad [B \subseteq A]$, thus $\color{tomato}{\neg}[A = B] \quad  \equiv \quad A\subsetneq B \quad \vee \quad B \subsetneq A$.
WLOG, say $A \subsetneq B$. Then $A$ must have fewer elements than $B$. Thus $|A| \neq |B|$.
Supplementary dated Mar 6 2014 : Does my writing above prove the contents of amWhy's sterling answer? Or are there better manners to prove it? 
 A: If we are given that $A, B$ are finite sets such that $|A| = |B|$, and if we know that $A \subseteq B$ or $B \subseteq A$, then we can conclude $A = B$.
Usually to prove that $A = B$, we need to prove that both $A\subseteq B$ AND $B\subseteq A$.
In this case, given finite $A, B$ with $|A| = |B|$, it suffices to show only one of the containments.
Note that the finite condition is necessary, the above isn't true of infinite sets. For example, take $A = \mathbb Z$, $B = \mathbb Q$.  Then both sets have the same cardinality, and $\mathbb Z \subseteq \mathbb Q$, but clearly $\mathbb Z \neq \mathbb Q$.
ADDED: To address the amended post, I think the contrapositive prefaced "for another way to construe this, ..." is the most succinct way to drive this relationship home. 
Indeed, if $A$ and $B$ are both finite sets and $(A \subsetneq B) \text{ or } (B\subsetneq A)$, then we know by definition of a proper subset that $$\Big((\forall a \in A (a \in B)) \land (\exists b \in B (b\notin A))\Big) \lor \Big((\forall b \in B(b \in A)) \land (\exists a \in A(a \notin B))\Big)$$ 
Therefore it is necessarily the case that $$|A| \lt |B|,\quad\text{or}\quad |B|\lt |A|$$
In either case, it follows that $|A|\neq |B|$.
A: Just for giving you another counter example, assuming you passed a good course about set theory: $$\text{card}{(\mathbb N_{\text{e}})}=\aleph_0=\text{card}{(\mathbb N)}$$ where $\mathbb N_{\text{e}}$ is the set of all even natural numbers.
