Independent systems and Lagrangians Definition 1:
The notion of independent systems has a precise meaning in probabilities. It states that the (joint) probability or finding the system ($S_1S_2$) in the configuration ($C_1C_2$) is equal to the probability of finding the system ($S_1$) in the configuration ($C_1$) times the probability of finding the system ($S_2$) in the configuration ($C_2$).
Definition 2:
However, in we consider fields systems, the practical tool is Lagrangians.
So I should say that 2 systems are independent if : 
$$ Lagrangian (S_1S_2) = Lagrangian (S_1) + Lagrangian (S_2)$$
The question:
Now, what is the relationship between these 2 definitions?
They could be only compatible, or they could be equivalent in the field domain.
Is there a way to "demonstrate" the latter from the former ?
 A: How about path integrals? The probability that a system evolves between state $|\phi_1\rangle$ and $|\phi_2\rangle$ is
$$\langle \phi_2|\phi_1 \rangle =\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)$$
where the measure $\mathcal{D}\phi$ is suitably defined and the action $S(\phi)$ is the integral of the Lagrangian (over whatever the physical coordinates are). 
Consider two systems, described by states $|\phi\rangle$ and $|\psi\rangle$, which are independent. Then the action is 
$$S(\phi,\psi)=S(\phi)+S(\psi)$$
And the probability of evolving between two configurations is 
\begin{align}
\langle\phi_2,\psi_2|\phi_1,\psi_1\rangle&=\int_{(\phi_1,\psi_1)}^{(\phi_2,\psi_2)}\mathcal{D}\phi\mathcal{D}\psi \exp \left(\frac{i}{\hbar}(S(\phi)+S(\psi))\right)\\
&=\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)\int_{\psi_1}^{\psi_2}\mathcal{D}\psi \exp \left(\frac{i}{\hbar}S(\psi)\right)\\
&=\langle \phi_2|\phi_1\rangle \langle \psi_2|\psi_2\rangle\\
\end{align}
So the probability is a product if the systems are independent. I picked specific states here but take $|\phi_i\rangle$ describing a system $S_i$ in configuration $C_{i1}$ and I think this gets what you want.
A: Well, you are not specifying the kind of probabilities you are talking about. Therefore, I refer to the tags and assume that we are talking about quantum (not statistical) probabilities. (See the edit at the end)
Then, I have to note that your probabilistic definition of independece does not make much sense.
Edit
Well, according to the comments, I should state that my point is that these definitions are different in their meaning. The first is the statistical independence (and should be refined for the case of QM) and the second is the absence of interaction. The following should be understood as a reference to quantum entanglement as a fact that non-interacting systems can be correlated.
End of edit
First of all, we usually believe that if we have the Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ of two systems, then the Hilbert space for the compound system is $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$. This is regardless of the interaction between the systems.
Next, when we talk about the probability of finding the system in state $|a\rangle$ while it is in state $|b\rangle$ (all normalized) we usually mean $|\langle a|b \rangle|^2$. Now, consider the following state in $\mathcal{H}$:
$$
|a\rangle=\frac{1}{\sqrt{2}}|1\rangle\otimes|\alpha\rangle+\frac{1}{\sqrt{2}}|2\rangle\otimes|\beta\rangle
$$
There is $0$ probability to find such system in the state $|1\rangle\otimes|\beta\rangle$ while there is $0.5$ prabability to find the first system in state $|1\rangle$ and $0.5$ probability to find the second system in the state $|\beta\rangle$. And such a state exists regardless of the interactions between the systems. See quantum entanglement.
Also, we can consider the state
$$
|a\rangle=|1\rangle\otimes|\alpha\rangle
$$
for which your rule more or less works.
Finally, we see that there are states of non-interacting systems such that your probabilistic equality does not hold, as well as there are states of interacting systems such that your equality holds. So, there is no connection between this equality and independence of the systems.
What is true, however, that the transition amplitudes (that is, matrix elements of $\exp(-iHt/\hbar)$) are indeed multiplicative for independent systems. That is because
$$
\exp(-iHt)=\exp(-it(H_1\otimes 1+1\otimes H_2))=\exp(-it H_1\otimes 1)\exp(-it1\otimes H_2)\\
=\exp(-iH_1t)\otimes\exp(-iH_2t)
$$
acts on the components of tensor product independently (also see levitopher's answer for path integral pespective).
What about the converse? We cannot say that any matrix element for the whole system is the product of the matrix elements for the separate systems because a priori we dont have the latter. We have to say something different, but I am now not sure how to state it conveniently.
Edit The tags have changed a bit, but still we can construct density matrices (and pretend that they correspond to some non-equilibrium situations) with similar properties. I think that the pure-state density matrix is ok.
For equilibrium density matrices, namely $\rho=\exp(-\beta H)$ it is true that probablilities mulptily for independent systems by the same argument as for the evolution operator. The same can be done via Euclidian path integrals.
As for the converse, I do not know the answer at the moment.
