Is horizontal composition of 2-morphism always unique? In $Cat$ seen as a 2-category, we have that horizontal composition of natural transformations is unique because the two "paths" in which the composition con be realised, turn out to be equal. 
But, in a general 2-category, can the horizontal composition be not unique? I.e. The two paths turn out to be different. I mean, taking the definition of strict or weak 2-categories, i don't see directly any requirements of this unicity. 
 A: In the definition of 2-category (and bicategory) that you are familiar with, the horizontal composition is defined as a functor $\mathcal{C}(B,C) \times \mathcal{C}(A,B) \rightarrow \mathcal{C}(A,C)$ and therefore the result of applying this functor to a pair of $2$-cells (one from each hom-category) is unique just because it is the application of a functor to a $1$-cell in a category. Applying a functor always gives a unique result.
However, you do bring up an interesting point. If you study weak higher category theory further, progressing from bicategories to tricategories (to tetracategories), you'll see why that explicit construction isn't always easy to deal with. As a result, category theorists have come up with many other, less explicit, models of weak higher categories. And in some of those models, you can interpret a given cell as being a composite of two other cells, but not the unique composite of those cells.
A: Let juxtaposition (or $\circ$) denote horizontal product and $\cdot$ denote vertical product.
I think what you have in mind is constructing the horizontal product of two $2$-morphisms $\eta \mu$ in in terms of the horizontal product of a $2$-morphism with a $1$-morphism:
$$\begin{align}
 \eta \mu &= (1_A \cdot \eta) (\mu \cdot 1_D) = (A \mu) \cdot (\eta D)
\\&= (\eta \cdot 1_B) (1_C \cdot \mu) = (\eta C) \cdot (B \mu)
\end{align}$$
which diagrammatically corresponds to the two different decompositions
$$ \begin{matrix}
\mathcal{C}  &\xrightarrow{A}& \mathcal{D} &\xrightarrow{C}& \mathcal{E}
\\  \| & \eta & \| & \mu & \|
\\ \mathcal{C}  &\xrightarrow{B}& \mathcal{D} &\xrightarrow{D}& \mathcal{E}
\end{matrix} 
\qquad = \qquad
 \begin{matrix}
\mathcal{C}  &\xrightarrow{A}& \mathcal{D} &\xrightarrow{C}& \mathcal{E}
\\  \| & 1_A & \| & \mu & \|
\\ \mathcal{C}  &\xrightarrow{A}& \mathcal{D} &\xrightarrow{D}& \mathcal{E}
\\  \| & \eta & \| & 1_D & \|
\\ \mathcal{C}  &\xrightarrow[B]{}& \mathcal{D} &\xrightarrow[D]{}& \mathcal{E}
\end{matrix} 
\qquad = \qquad
 \begin{matrix}
\mathcal{C}  &\xrightarrow{A}& \mathcal{D} &\xrightarrow{C}& \mathcal{E}
\\  \| & \eta & \| & 1_C & \|
\\ \mathcal{C}  &\xrightarrow{B}& \mathcal{D} &\xrightarrow{C}& \mathcal{E}
\\  \| & 1_B & \| & \mu & \|
\\ \mathcal{C}  &\xrightarrow[B]{}& \mathcal{D} &\xrightarrow[D]{}& \mathcal{E}
\end{matrix}
$$
where I'm lazy and put my diagrams and algebra in the same ordering, rather than the usual reverse convention. Note that I draw squares not only because it's easier to draw, but because I find it easier to to do algebra with squares than with than the usual globular picture of $2$-morphisms as disks.

Normally, the definition of bicategory bypasses this completely: the horizontal composition is a given as part of the bicategory structure. Specifically, that horizontal composition is a functor between the two categories $M:\hom(\mathcal{C}, \mathcal{D}) \times\hom(\mathcal{D}, \mathcal{E}) \to \hom(\mathcal{C}, \mathcal{E})$.
However, it is not hard to show that this implies that any bicategory satisfies the interchange law:
$$ M(\alpha \cdot \beta, \gamma \cdot \delta) = M\left( (\alpha, \gamma) \cdot (\beta, \delta) \right)= M(\alpha, \gamma) \cdot M(\beta, \delta) $$
or, in the usual infix notation,
$$ (\alpha \cdot \beta)(\gamma \cdot \delta) = (\alpha \gamma) \cdot (\beta \delta) $$
So, the horizontal product can indeed be expressed in terms of products with identity $2$-morphisms, as in the argument above.

However, this phenomenon is sort of an artifact of the fact that bicategories only have "two-dimensional" structure — in general, the coherence condition we want imply that the three diagrams above are all equivalent, and uniquely so (up to equivalence).
But since there are no $3$-morphisms in a bicategory, equivalent means equal for $2$-morphisms.
