Contrasting definitions of bimodules? An illusion? Recently my definition of a bimodule over a $k$-algebra has been challenged and I believe both definitions to be equivalent, am I wrong? 

Notation: $k$ is a commutative ring and $A$ is a (unital associative) $k$-algebra.  

Definition 1:
An $(A,A)$-bimodule $M$ is an abelian group together with a left and right $A$-module structure such that: \begin{equation}
(\forall m \in M) (\forall a,b \in A) (a \cdot m) \cdot b = a \cdot (m \cdot b)
\end{equation}

Definition 2:
An $(A,A)$-bimodule $M$ is an $k$-module together with a left and right $A$-module structure such that: \begin{equation}
(\forall m \in M) (\forall a,b \in A) (a \cdot m) \cdot b = a \cdot (m \cdot b)
\end{equation}

Reasoning
$A$ is a $k$-algebra then the inclusion of $k$ into $A$ induces an action of $k$ on $M$; whence $M$ must also be a $k$-module if it satisfies definition $1$, conversely if $M$ satisfies definition $2$ then it satisfies definition 1 by definition of a $k$-module.  

Apparently my logic is flawed but I don't see why...
 A: The first definition is what you might call a "ring" bimodule. It only asks for $A$ to be a ring, and it doesn't mind if $A$ has any $k$ algebra structure.
I would call your second definition an algebra-bimodule, but I think you're missing an axiom. Not only does $M$ have a $k$-module structure, but this structure is compatible with $A$'s $k$-structure! You should also insist that $(ka)\cdot m=a\cdot (km)=k(a\cdot m)$ and $m\cdot(kb)=(km)\cdot b=k(m\cdot b)$ for all choices of $k,a,b,m$. This makes sure that the $k$-module action on $M$ is in sync with the $k$-structure on $A$.
In the first definition, simply no comment is made about the $k$-algebra structure on $A$ in relationship to $M$. Certainly you can induce a $k$ module structure on $M$ using the $A$ action on $M$, but this is just not mentioned.
Finally, you could also remember that all rings are $\Bbb Z$ algebras, and if you're working with unital modules (as most people do) then you are automatically guaranteeing that every ring-bimodule is also a $\Bbb Z$-algebra bimodule.
A: In Definition 2 the action of $k$ on $M$ induced by the action of $k$ on $A$ can be different from the given $k$-action on $M$.
A: As you say in your reasoning, the $A$-module structure on $M$ induces a $k$-module structure. So does the $B$-module structure. But without added conditions these two $k$-module structures could be different.
For example, take $A=B=M=\mathbb{C}$ with the bimodule structure
$$a\cdot m\cdot b=am\overline{b},$$
where $\overline{b}$ is the complex conjugate of $b$.
This is a perfectly good bimodule structure if you're dealing with abstract rings (or $\mathbb{R}$-algebras), but working over $\mathbb{C}$, the standard definition would require that (in your first definition) the $k$-module structures on $M$ induced by the $A$-module and $B$-module structures are the same, or equivalently that (in your second definition) the $A$-module and $B$-module structures are both $k$-bilinear.
