Lambert W of a constant multiple $ \ln(cx) $ can be expressed as $ f(c) + \ln(x) $, where $ f(c)=\ln(c) $.
Does the lambert W function have a similar property? (Can $ W(cx) $ be expressed as $ f(c)+W(x) $ for some function $ f $).
 A: Short answer: no. But here's what happens if you try (along one particular path of analogous reasoning).
Every identity about a function is equivalent to an identity about its inverse. For the logarithm,
$$
\ln(x\,y) = \ln x + \ln y 
\label{1}\tag{1}$$
is equivalent to
$$
\exp X \, \exp Y = \exp(X+Y), 
$$
which you can see by setting $x = \exp X$ and $y = \exp Y$. (I'm suppressing domain considerations for concision.)

Now, Lambert's $W$ function is the inverse of the product-exponential, i.e.
$$
X = W(x) 
$$
means that
$$
x = X \, \exp X. 
$$
(Again, we should be careful of domains, but we're trying get intuition here first.) This is a more complicated relationship, so we should expect any identity to be more complicated than the analogous one for the logarithm.
Here's a derivation of one possible identity that involves a sum of $W$ values:
\begin{align}
(X + Y) \, \exp(X+Y) &= (X+Y) \, ( \exp X \, \exp Y ) \\ 
&= (X \exp X) \, \exp Y + \exp X \, (Y \exp Y) \\ 
&= (X \exp X) \, \frac{Y \, \exp Y}{Y} 
+ \frac{X \, \exp X}{X} \, (Y \exp Y) \\
&= (X \exp X) \, (Y \exp Y) \, \biggl( \frac{1}{X} + \frac{1}{Y} \biggr). 
\end{align}
Applying $W$ to both sides, where $X = W(x)$ and $Y = W(y)$ yields
$$
W(x) + W(y) 
= W \Biggl( xy \, \biggl( \frac{1}{W(x)} + \frac{1}{W(y)} \biggr) \Biggr). 
$$
If we try to mimic the left side of Equation $(\ref{1})$ instead, it's even worse:
$$
(X \, \exp X) \, (Y \, \exp Y) 
= X \, Y \, \exp(X + Y)
= \frac{XY}{X + Y} \, (X + Y) \, \exp(X + Y), 
$$
but this doesn't lead anywhere useful that I can see. Hence, this shows that $W(xy)$ is equal to a mess.
