Here is a solution sketch. Unfortunately, I don't have enough reputation to embed pictures yet, nor can I embed more than two links.

As illustrated in the question, we refer to the segments as red, blue, or green, depending on their directions. In what follows, "segment" always refers to a segment appearing in a given covering of the plane, subject to the requirements of the question.

Define a vertex to be a point at which two segments (of different directions) meet. A vertex falls into exactly one of four types, depending on which kinds of segments are adjacent to it: red-blue, red-green, blue-green, red-blue-green. We propose to join the vertices with auxiliary edges, so as to realize the parallelograms illustrated in the question.

To this end, consider (for example) a vertex $v$ incident with blue and red (and possibly also green) edges, as shown. The unit-distance constraint implies that a blue edge and a red edge lie on the grey lines shown. The no-60-degrees constraint implies that these two edges fall into one of the two configurations on the right. In either case, we find a new vertex $v'$ where the two new segments meet. Join $v$ and $v'$ with an auxiliary edge. Repeat this process for all vertices $v$.

Image 1

In the first case, the resulting auxiliary edge $\overline{vv'}$ can't meet any other auxiliary edge in its interior. There will be exactly one more auxiliary edge incident with $v'$, going up and to the left. There may be one or two auxiliary edges incident with $v$, depending on whether $v$ is incident with a green segment. If so, the three auxiliary edges incident with $v$ are at 120-degree angles with each other. If not, then $v$ is incident with exactly one auxiliary edge which is collinear with $\overline{vv'}$.

However, in the second case, the conditions force the appearance of a small hexagon (see below), and three auxiliary edges will meet at their midpoints, i.e. the center of the small hexagon.

Image 2

Consider the union of all the auxiliary edges. From the last two paragraphs, we know that the auxiliary edges are characterized by these local neighborhoods:

1. A single line (along which segments of exactly two different colors meet).
2. Three segments meeting at 120-degree angles (where the intersection point is a vertex incident with segments of all three colors).
3. Six segments meeting at 60-degree angles (where the intersection point is the center of a small hexagon).

The auxiliary edges partition the plane into (possibly infinite) regions, bounded by polygonal curves all of whose sides are at 60 or 120 degree angles to one another. Since no vertex lies in the interior of such a region, all segments inside one region are the same direction (i.e. color).

By the description of 2. and 3. above, each 60-degree angle (resp. 120-degree angle) of the boundary of a region determines the color of the segments inside the region. It is now straightforward to check that the only consistent possibilites for the shape of a region are as follows:

• parallelogram (angles: two 60's, two 120's)
• infinite half-strip (angles: one 60, one 120)
• infinite 60-degree sector (angles: one 60)
• infinite 120-degree sector (angles: one 120)
• infinite strip (no angles)
• half-plane (no angles)

Furthermore, given the shape and orientation of one of the first four types of regions (those with vertices, hence angles), the color of the segments inside it is uniquely determined by the fact that these segments must not be parallel to (any part of) the boundary of the region.

Each side of a region must either be infinite or $(2k+1)/2 \cdot \sqrt{3} / 2$, for a nonnegative integer $k$. Conversely, any decomposition of the plane into regions (of the kinds described in the last paragraph, and subject to constraints 1. 2. 3. above), whose side lengths are either infinite or $(2k+1)/2 \cdot \sqrt{3} / 2$, comes from a desired covering of the plane, which can be easily drawn. Furthermore, the resulting covering of the plane is unique, except in the case where the only regions that appear are infinite strips and half-planes, in which case one needs to specify the color of the segments appearing in one of the regions (two choices).

The previous paragraph implies that no finite configuration can force a periodic covering of the plane.

The most interesting plane coverings arise when every region is finite. In this case, the abstract arrangement of regions is given by the picture in the question statement. The side lengths can be arbitrary chosen, subject to the constraint that opposite sides of paralellograms have the same length. So it is exactly enough to specify three infinite (two-sided) sequences of nonnegative integers, corresponding to $k$ above. Each sequence corresponds to the lengths of the sides running in one of the three directions. In the case where all these integers are zero, we obtain the tiling of the plane by small hexagons.