I'm trying to get a full grasp on the relationship between many of the units that are used in kinetic physics. I've found that it is possible to make a venn-diagram that shows the factors of many of the units: Impulse in the centre, Force and Velocity between two circles each and Mass, Acceleration and Time on the Outside. The only space in the diagram that is missing is mass multiplied by time.
Is there a such thing as a kilogram-second? Is it used for anything in physics?
There's nothing too common or universal, especially in basic kinetics, and moreover this shouldn't be much of a surprise. There are infinitely many unit combinations one can construct, and there are only finitely many combinations we bother to give special names to, so necessarily some combinations will not have special names.
You might object that maybe at least the simple combinations should have some use, but in fact the definition of "simple" you've implicitly employed is rather arbitrary. Why use mass $M$, acceleration $A$, and length $L$? Why not swap out acceleration for time $T$, for example? At least this would coincide with the SI base units, which somebody, somewhere decided were pretty decent choices for a base.
Let's see what happens when everything is written in terms of $M$, $L$, and $T$.
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline & 1 & M & A & T & MA(\text=F) & MT(=?) & AT(\text=V) & MAT(\text=P)\\ \hline M & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1\\ \hline L & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1\\ \hline T & 0 & 0 & -2 & 1 & -2 & 1 & -1 & -1\\ \hline \end{array}$$
The top row are the eight regions of your Venn diagram: dimensionless, mass, acceleration, time, force, unknown, velocity, and momentum. Each column shows the decomposition into $M$, $L$, and $T$. For example, $MAT = P = M^1 L^1 T^{-1}$.
But in my choice of $M$, $L$, and $T$, there are "simpler" combinations (those with smaller numbers) than some given that are missing. If you let the powers on $M$ and $L$ be $0$ or $1$, and you let the power on $T$ range from $-2$ to $+1$, then in fact there are another eight combinations we can come up with:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline & 1/T^2 & 1/T(\text=f) & AT^2(\text=L) & AT^3 & M/T^2 & M/T & MAT^2 &MAT^3\\ \hline M & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ \hline L & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1\\ \hline T & -2 & -1 & 0 & 1 & -2 & -1 & 0 & 1\\ \hline \end{array}$$
We happen to have special names for the inverse of time (frequency $f$) and for acceleration times the square of time (length $L$). But many of these combinations also have no particular need for a separate name. And we could keep going, allowing for higher (and more negative) powers of our units, getting even more esoteric combinations, most of which won't have special names.
There's no particularly deep reason for any of this. We simply name quantities that show up often in our favorite formulas.
Two examples of the product of mass and duration being quantified over vastly different ranges:
1) It is not entirely uncommon to cost a truck on a tonne.hr basis. Here the payload is measured in tonnes and the trip duration in hours. The example given is:
"if a truck hauls 35 Tonnes and the trip time is 4 hours, the number of tonne$\cdot$hours is 140 (35 x 4). If the standard cost is USD 3.00 per tonne$\cdot$hour, the truck cost would be USD 420 (= 140 x USD 3.00)."
in SI units a tonne$\cdot$hour would equate to 3.6 million kg$\cdot$seconds.
2) In physics, the product of mass and time appear in the phenomenon "Zitterbewegung". Zitterbewegung provides a physical interpretation for the complex phase factor in the Dirac wave function. It is manifest as a rapid motion of the position of elementary particles that obey the Dirac equation. This motion repeats with a frequency $2mc^2/h$. Therefore, over a timespan $\delta t$, a Dirac particle with mass $m$ will undergo $m \delta t/(h/2c^2)$ 'Zitter cycles'. (It's no coincidence that in essence the same constant $h/2c^2$ measured in SI units kg$\cdot$s appears here as well as in Void's answer.) In other words, when measured in natural units, the product of mass and time equates to the number of 'double Zitter cycles' a Dirac particle will undergo.
This is a question of units, and I am not going to pretend I present an answer which would be anyhow natural from the point of view of classical non-relativistic physics. But consider the following. We have a version of the Heisenberg uncertainty principle $$\Delta E \Delta t \geq \frac{\hbar}{2}$$ This for example tells us that an excited state with a finite lifetime will have a minimal indefiniteness of energy. In nuclear physics this uncertainty is connected with uncertainty of mass through the $E = mc^2$ relation, so we have $$\Delta m \Delta t \geq \frac{\hbar}{2 c^2}$$ That is, in SI units, the mass-time uncertainty has the dimensions $\rm kg \cdot s$. The mass of a large number of excited nuclei can be indeed measured by weighing (even though I doubt that for most cases they can be weighed quickly enough) so the mass uncertainty has a direct physical meaning as has the lifetime.
The stated mass-time uncertainty obviously applies universally, but nuclear physics is the place where it is experimentally relevant in the sense of "weight". E.g. in particle physics this relation also holds, but they never really "weigh" particles but find out their mass/energy from collisions, so we would be more inclined to call this an energy-time or even momentum-time uncertainty since the rest mass of the particle is basically overwhelmed by momentum. Nonetheless, the relation is e.g. used to constrain the resonance width.
Now for some arbitrary magic with units. To complete your Venn diagram, you could state that mass actually has units of Joules and the conversion is provided by multiplying by the square of light-speed ($E=mc^2$). The intersection of $m$ and $t$ would thus be $S$, the action, for which we have $[S]=[\hbar]=\rm J \cdot s$. The connection with the previous example is clear from the fact that quantum mechanics actually says that there is a physically significant minimal quantum of action in nature proportional to $\hbar$.
Your Venn diagram would look something like this:
But once you start using $c$ to decimate your units, you do it consistently. Namely, you should also make time the same units as space and thus $[v]=1$ and $[a]=[\rm length]^{-1}$. But this isn't so horrible, because you then have $[F] = [\rm Energy][length]^{-1}$ which makes total sense and all the central stuff has units $[\rm Energy]$ which also isn't way so bad.
An example that comes from Compton scattering would be something like
$m_e /\lambda$
would have $kg \cdot m$, $1/\lambda$ is related to the momentum of the photon, but not $kg\cdot s$.
By itself this quantity is not usefull, but it does appear during the calcuations
Edit: I gave a bad example, edits above
