Consider the logistic map,
$x_n = r \: x_{n-1} \left( 1 - x_{n-1} \right)$
If we generalize this to a complex function $f : \mathbb{C} \mapsto \mathbb{C}$, we get,
$f \left( z \right) = r \: f \left( z-1 \right) \left( 1- f \left( z-1 \right)\right)$
$$f \left( z \right) = r \left( f \left( z-1 \right)- \left( f \left( z-1 \right) \right)^2 \right) \: \dots \: \left( 1 \right) $$
Assume that there exists a function $u: \mathbb{C} \times \mathbb{R} \mapsto \mathbb{C} \times \mathbb{R} $, such that,
$\displaystyle{f \left( z \right) = \mathcal{L} \{ u \left( z, t \right) \} \left( s \right) = \int_0^\infty u \left( z, t\right) e^{-st} dt }$
As per integration by parts,
$$\mathcal{L} \{ u \left( z, t \right) \} \left( s \right) = \frac{1}{s} \left( u \left( z, 0 \right) + \mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( s \right)\right) \: \dots \: \left( 2 \right)$$
Where $\dot{u} \left( z, t\right) = \frac{\partial u \left( z,t \right)}{\partial t}$
Notice that equations $\left( 1 \right)$ and $\left( 2 \right)$ have the same L.H.S. They are identical in form if we take,
$s = \frac{1}{r}, \\ u \left( z, 0 \right) = f \left( z-1 \right), \\ \mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( s \right) = - \left( f \left( z-1 \right) \right) ^2 \text{, or,}$
$\mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( s \right) = - \left( u \left( z, 0 \right) \right)^2$
Replacing $s$ with $\frac{1}{r}$,
$$ \mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( \frac{1}{r} \right) = - \left( u \left( z, 0 \right) \right)^2 \dots \left( 3 \right) $$
The same equation may be rewritten in integral form,
$\displaystyle{ \left( u \left( z,0 \right) \right)^2 + \int_0^\infty \frac{\partial u \left( z,t \right)}{\partial t} e^{-\frac{t}{r}} dt = 0 }$
Applying the Leibniz integral rule,
$\displaystyle{ \left( u \left( z,0 \right) \right)^2 + \frac{d}{dt} \int_0^\infty u \left( z,t \right) e^{-\frac{t}{r}} dt = 0 }$, or,
$\displaystyle{ \left( u \left( z,0 \right) \right)^2 + \frac{d}{dt} f \left( z \right) = 0 }$
But $\frac{d}{dt} f \left( z \right) = 0$, so,
$ \left( u \left( z,0 \right) \right)^2 = 0 \implies u \left( z,0 \right) = 0 $
$$ u \left( z,0 \right) =0 \: \dots \: \left( 4 \right) $$
But something is wrong in the above derivations because according to the above result, $f \left( z-1 \right) = u \left( z,0 \right) = 0$ which yields a constant function.
This means that our assumptions in the beginning were wrong. However, is there a way to proceed with the problem (of expressing logistic maps non-recursively) along similar lines of thinking?
Taking a look at the assumption $ f ( z ) = \mathcal L _ t \{ u ( z , t ) \} ( s ) $, one can see that $ s $ does not appear on the left-hand side of the equation, which means that the Laplace transform of $ u $ is assumed to be constant (when talking about a fixed $ z $). It is well-known that no such function exists, and you need to generalize the notion of functions to find candidates for $ u $. Doing that, it is also well-known that $ u $ must be a multiple of Dirac $ \delta $ function. More precisely, you must have $ u ( z , t ) = f ( z ) \delta ( t ) $.
The point is that generalized functions do not exactly act like functions. Dirac $ \delta $ function, for example, on one hand acts like a function that that take the value zero at any point other than the origin, but on the other hand, has positive integral; for any $ \epsilon > 0 $ you have $ \int _ { - \infty } ^ { - \epsilon } \delta ( t ) \ \mathrm d t = \int _ \epsilon ^ { + \infty } \delta ( t ) \ \mathrm d t = 0 $, while $ \int _ { - \infty } ^ { + \infty } \delta ( t ) \ \mathrm d t = 1 $. This can't happen for any genuine function, even if you allow the function to take the value $ + \infty $; considering a function $ f : ( - \infty , + \infty ) \to [ - \infty , + \infty ] $ such that $ f ( x ) = 0 $ for any $ x \in ( - \infty , 0 ) \cup ( 0 , + \infty ) $ and $ f ( 0 ) = + \infty $, you still have $ \int _ { - \infty } ^ { + \infty } f ( t ) \ \mathrm d t = 0 $, in the sense of Lebesgue integration.
That said, it should be obvious that you have to avoid treating $ u $ like a function that takes definite values at given points $ z $ and $ t $. As long as you're keeping a distance from $ t = 0 $ though, it acts like a constant zero function, which may explain why you ended up with $ u ( z , 0 ) = 0 $: because $ u ( z , \epsilon ) $ behaves like the constant zero function for $ \epsilon \ne 0 $, even for arbitrarily small values of $ \epsilon $. But note that this explanation is rather loose, as $ u ( z , 0 ) $ has no meaning, because Dirac $ \delta $ function has no definite value at the origin.