对于scipy 1.7.1，已经实现了新的功能，并且dendogram函数在输出字典中还返回一个条目“leaves_color_list”，可以用来轻松执行此任务。
下面是OP的工作代码（见最后一行“新代码”）

import numpy as np
import matplotlib.pyplot as plt
from scipy.cluster.hierarchy import linkage, dendrogram

# DATA EXAMPLE
x = np.array([[ 5, 3],
              [10,15],
              [15,12],
              [24,10],
              [30,30],
              [85,70],
              [71,80]])

# DENDROGRAM
plt.figure()
plt.subplot(121)
z = linkage(x, 'single')
d = dendrogram(z)

# COLORED PLOT
# This is what I would like to achieve. Colors are assigned manually by looking
# at the dendrogram, because I failed to get it from d['color_list'] (it refers 
# to links, not observations)
plt.subplot(122)

#NEW CODE
plt.scatter(x[d['leaves'],0],x[d['leaves'],1], color=d['leaves_color_list'])

以下方法似乎有效。树状图返回的字典包含'color_list'和链接的颜色。而'icoord'和'dcoord'分别与x，y，绘制这些连杆的坐标。这些X位置是5、15、25、.当链接从一点开始时。因此，测试这些x位置可以将我们从连杆带回到相应的点。并允许将链接的颜色指定给该点。

import numpy as np
import matplotlib.pyplot as plt
from scipy.cluster.hierarchy import linkage, dendrogram

# DATA EXAMPLE
x = np.random.uniform(0, 10, (20, 2))

# DENDROGRAM
plt.figure()
plt.subplot(121)
z = linkage(x, 'single')
d = dendrogram(z)
plt.yticks([])

# COLORED PLOT
plt.subplot(122)
points = d['leaves']
colors = ['none'] * len(points)
for xs, c in zip(d['icoord'], d['color_list']):
    for xi in xs:
        if xi % 10 == 5:
            colors[(int(xi)-5) // 10] = c
for point, color in zip(points, colors):
    plt.plot(x[point, 0], x[point, 1], 'o', color=color)
    plt.text(x[point, 0], x[point, 1], f' {point}')
plt.show()

PS：This post关于匹配点与它们的聚类也可能是相关的。