Shapley values without intercept (or without `expected_value`)
This is very similar to fitting a linear regression and not including an intercept, and I think they will face similar issues.
To be very concrete, consider an example with $f(x)=1,\ E=1, \ \phi_1=1, \ \phi_2=-1$.
Then your scaling factor is undefined, trying to divide by zero.  Well OK, but you won't often get such exact numbers.  Let's tweak them to
$$f(x)=1,\ E=1, \ \phi_1=1.01, \ \phi_2=-1.$$
Now the scaling factor is $1+1/.01=101,$ so that $\hat{\phi}_1=102.01$ and $\hat{\phi}_2=-101$.  This isn't locally too bad, because it still indicates the relative importance of the two variables in making this prediction.  But if you want to compare this to other points $x$ (where say $f(x)=2,\ \phi_1=1,\ \phi_2=0$, so that $\hat{\phi}_1=2,\ \hat{\phi}_2=0$) things will look quite strange, and aggregating importances over your entire sample may be very misleading.
For the connection to linear regression and more, see this excellent description of shapley values.