After reading the papers by Rufus Isaacs [1] and George Spencer-Brown [2], I have reached to the conclusion that spiral chain edge coloring algorithm [3] gives answer to the question in affirmative.
[1] Rufus Isaacs, "Infinite families of nontrivial trivalent graphs which are not tait colorable", American Math Monthly 73 (1975) 221-239.
[2] George Spencer-Brown, "Uncolorable trivalent graphs", Thirteenth European Meeting on Cybernetics and Systems Research, University of Vienna, April 10, 1996.
[3] I. Cahit, Spiral Chains: The Proofs of Tait's and Tutte's Three-Edge-Coloring Conjectures. arXiv preprint, math CO/0507127 v1, July 6, 2005.
Let $G$ be a bridgless planar $3$-regular graph and suppose it is given a plane embedding. Then by the four-colour theorem, there exists a proper $4$-face-colouring of $G$, say using $(1,2,3,4)$.
Now notice that since $G$ is bridge-less, all edges see two distinct faces which have different colours.
Colour the edges which see faces coloured $1,2$ or $3,4$ blue. Colour edges which see faces coloured $1,3$ or $2,4$ red. Colour the edges which see faces coloured $1,4$ or $2,3$. Note this colours all edges in our graph as every edge sees two distinct faces.
Since $G$ is $3$-regular, this colouring is a $3$-edge-colouring of $G$. If two adjacent edges were given the same colour, then either they both see the same coloured faces, or they each see two different faces.
In the case where both see the same coloured faces, this implies $G$ has a vertex of degree $2$. In the other case where the edges see all different faces, that implies that $G$ has a vertex with degree greater than $3$.
Best Answer
The four colors can be represented by bit-vectors of length 2.
We can define $3$ transformations, $L, R, B$ on these bit-vectors, where $L$ inverts the left-bit, $R$ inverts the right-bit, and $B$ inverts both bits. Each of these is bijective with no fixed point.
Note that composing all three of these gives the identity, and composing any two of them gives the third.
Now consider a maximal planar graph. If the vertices can be $4$-colored then each edge determines one of these transformations and the edges of a triangle must have all $3$ transformations in order to compose to the identity transformation.
Conversely, if the edges can be $3$-colored such that the $3$ edges of each triangle have different colors, then $L,R,B$ can be used for the colors, and the transformations can be used to extend a color on a starting vertex to the whole graph.