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...

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.
