Function to determine path from imported module output - snakemake

Question

We have a project in which we combine multiple independent Snakemake workflows using subworkflow. Now I'm trying to convert the project to use module instead but I'm facing difficulties due to either lacking features or documentation.
Previsouly we could in the "parent" Snakefile include a subworkflow like this (from the documentation:
subworkflow otherworkflow:
workdir:
"../path/to/otherworkflow"
snakefile:
"../path/to/otherworkflow/Snakefile"
configfile:
"path/to/custom_configfile.yaml"
rule a:
input:
otherworkflow("test.txt")
output: ...
shell: ...
and then e.g. indicate that files should be found in the workdir of the otherworkflowby specifying otherworkflow(<output_file>).
Problem:
When using Snakemake module an equivalent options seems to be missing.
My current workaround:
I reference the output of the module via rules.<...>.output, e.g.:
Snakefile 1:
rule a:
output: "some_file.txt"
shell: ...
Snakefile 2:
module other_workflow:
snakefile: "<path to Snakefile 1>
workdir: "<some workdir path>"
use rule * from otherworkflow as other_*
rule b:
input: rules.other_a.output[0]
shell: ...
Which works to some extent but is not as nice to read as the previous syntax from subworkflow.
Also: if the output contains a wildcard, then this workaround does not work anymore. (expand-ing the wildcard e.g. expand(rules.other_a.output[0], wildcard_name=<value>) does not work; what does work is format-ing rules.other_a.output[0].format(wildcard_name=<value>))
I cannot find a similar functionality when using a Snakemake module for resolving filenames relative a modules workdir.
Q: Is there any solution for indicating an output is from an imported snakemake module and constructing the output path from workdir as we had with subworkflow?