Modules over associative algebras are just special cases of "ordinary" modules over rings? By module over a ring, I mean always a right-module. All rings are supposed to be unital, and the module fulfills $m\cdot 1 = m$. If $R$ is commutative and $M$ a right-module, we can define $rx := xr$ and get also a left-module. In this situation we call $M$ a bimodule, and if not otherwise said this natural definition is applied if I speak of bimodules.
An algebra $A$ over a commutative ring $R$ is itself a ring with $1$, which is also a $R$-module, and such that for all $r \in R$ and $x,y \in A$ we have
$$
 (rx)y = r(xy) = x(ry).
$$
As the ring is assumed to be commutative, we essentially have a bimodul over $R$ as written above.
A module $M$ over an algebra $A$ which is defined over some ring $R$ is itself an $R$-module, for which we have an operation $M \times A \to M$ such that
for each $u, v \in M$ and $x,y \in A$ and $r \in R$ we have
(1) $(u+v)x = ux + vx$
(2) $v(x+y) = vx + vy$
(3) $(vx)y = v(xy)$
(4) $v1 = v$
(5) $(rv)x = r(vx) = v(rx)$.
This is the standard definition I see everywhere. But I guess I observed the following. For an algebra we can embed $R$ into $A$ by identifying it with $R\cdot 1$ for $1 \in A$. So the assertion that $A$ must be an $R$-module is implied by (1), (2), (3) and (4). So we can just define a module over an algebra as an ordinary module over $A$ seen as a ring with $1$. As $A$ is in general not commutative, this is just a right-module. But by the embedding, and the additional requirement for an algebra, every element of $R$ is in the center $Z(A) = \{ x : xy = yx \mbox{ for all y} \in A \}$, restricted to $R$ everything is fine and we again have a bimodule. So the only thing that makes modules over algebras special here is (5). But here we have
\begin{align*}
 (rv)x & = (vr)x & \mbox{definition of $M$ as bimodul over $R$} \\
       & = v(rx) & \mbox{by (3)} \\
       & = v(xr) & \mbox{$R$ is central in $A$} \\
       & = (vx)r & \mbox{by (3)} \\
       & = r(vx) & \mbox{definition of $M$ as bimodul over $R$}
\end{align*}
so we see that in the third and last line we have recovered everything from (5).
So as I see it, a module over an algebra is just an ordinary module over the algebra seen as a ring, so why bother with these extra definition? I guess if we generalise further, for example look at non-associative algebras, then they are no longer rings and we cannot define modules over them as special cases of modules over rings. But most of the time (and in the textbooks I am reading right now) such generalizations are not considered, but nevertheless most of the time modules over algebras are defined separately to modules over rings.
So why that? Or have I overlooked something and my computations are wrong?
Remark: These modules over algebras come from the representation theory of (finite) groups.
 A: You need condition (5) because, in your definition of $A$-module, you asked for $M$ to be an $R$-module right at the start.  Condition (5) is there to ensure that the structure of $A$-module which you define on $M$ is compatible with the already existing $R$-module structure.
If, in your definition, you had simply asked for $M$ to be an abelian group, then condition (5) would be redundant, for the reasons you noted.  This is, for instance, what is done in Bourbaki's Algebra, chapter 3: if $A$ is an $R$-algebra, they define an $A$-module to be a module over the underlying (not necessarily commutative) ring of $A$.
A: Indeed your proof seems to work fine assuming in the definition you require $M$ to be a right $R$-module. In that case one could simplify the definition by requiring that a module of an algebra $A$ is just an $A$-module (the $R$-module structure can be derived in the way you showed).
On the other hand if we reguard $M$ as a left $R$-module things change.
If $M$ is just a left $R$-module the condition $(5)$ is required in order to ensure that the left action of $R$ and right action of $A$ (and so also the induced right action of $R$) are compatible: that is that $M$ is a $(R,A)$-module. 
Note that in general even for $(R,R)$-bimodules (i.e. left and right $R$-modules where the two actions are compatible) it is not required that the left and right actions coincide.
Hope this helps.
