Example Calculation
I realized I could make some progress on an open problem if I introduced a new operation: $f(*) x = f(x)$, where $f$ is a function and $x$ is a variable, then I could make progress. It seems initially just like some new notation but its not as trivial. Consider the following transformation:
$$ \left( \begin{array}{cc} x \\ y \end{array} \right)= % \left( \begin{array}{cc} e^{*} & 0 \\ 0 & e^{*} \end{array} \right) \cdot( \left( \begin{array}{cc} \ln(*) &\ln(\cos(*)) \\ \ln(*) & \ln(\sin(*)) \end{array} \right) \left( \begin{array}{cc} r \\ \theta \end{array} \right)) $$
Note: $$ \left( \begin{array}{cc} e^{*} & 0 \\ 0 & e^{*} \end{array} \right) \cdot \left( \begin{array}{cc} \ln(*) &\ln(\cos(*)) \\ \ln(*) & \ln(\sin(*)) \end{array} \right) \neq \left( \begin{array}{cc} * &\cos(*) \\ * & \sin(*) \end{array} \right)$$ As one gets the wrong answer then as it seems associative property of matrices breaks down. Why? Because if one assumes it doesn't then one gets the answer:
$ x = r + \cos \theta$ and $ y = r + \sin \theta$ whereas if one solves the bracketed matrices first one gets:
$ x = r \cos \theta$ and $ y = r \sin \theta$. But it seems we can multiply by inverses that takes precedence over other operations (inverse precedence conjecture):
Multiplying both sides by an inverse of the leftmost matrix on the R.H.S:
$$ \left( \begin{array}{cc} \ln(*) & 0 \\ 0 & \ln(*) \end{array} \right) \cdot \left( \begin{array}{cc} x \\ y \end{array} \right)= % \left( \begin{array}{cc} \ln(*) &\ln(\cos(*)) \\ \ln(*) & \ln(\sin(*)) \end{array} \right) \left( \begin{array}{cc} r \\ \theta \end{array} \right) $$
Once again, multiplying with an inverse (the inverse was using a clever guess in this case):
$$ (\left( \begin{array}{cc} 1 & 0 \\ 0 & \tan^{-1} e^* \end{array} \right) \cdot( \left( \begin{array}{cc} \exp(*)^2 & \exp(*)^2 \\ -1 & 1 \end{array} \right) \cdot (\left( \begin{array}{cc} \ln(*) & 0 \\ 0 & \ln(*) \end{array} \right) \cdot \left( \begin{array}{cc} x \\ y \end{array} \right))))= \left( \begin{array}{cc} r \\ \theta \end{array} \right) $$
Questions
Is there any case where the inverse precedence conjecture fails? Is there a general procedure find an inverse of a matrix with $*$ operators in it?
Lets take a 'general case' of an mxn matrix 'multiplying' a n element column vector, that is
$$ \left( \begin{array}{cccc} f_{11}(\star) & f_{12}(\star) & \ldots & f_{1n}(\star) \\ \vdots \\ f_{m1}(\star) & f_{m2}(\star) & \ldots & f_{mn}(\star) \end{array} \right) \left( \begin{array}{c} x_1\\ \vdots \\ x_n \end{array} \right) = \left( \begin{array}{cccc} f_{11}(x_1) + f_{12}(x_2) +\ldots + f_{1n}(x_n) \\ \vdots \\ f_{m1}(x_1) + f_{m2}(x_2) + \ldots + f_{mn}(x_n) \end{array} \right) $$
Thus we see that this notation is equivalent to a function $$ f \left(\left( \begin{array}{c} x_1\\ \vdots \\ x_n \end{array} \right) \right) = \left( \begin{array}{cccc} f_{11}(x_1) + f_{12}(x_2) +\ldots + f_{1n}(x_n) \\ \vdots \\ f_{m1}(x_1) + f_{m2}(x_2) + \ldots + f_{mn}(x_n) \end{array} \right) $$ and we are considering functions of column vector expressable as a sum of functions of each element, the matrix notation being useful as book-keeping.
To obtain an inverse function of $f$ is not possible in general. However, if $m=n$ and the function-matrix has one non-zero element per row and collumn, so only $f_{1\sigma(1)},\ldots,f_{n\sigma(n)}$ are non zero (and invertable), where $\sigma$ is a permutation function on $1,\ldots,n$, then
$$ f \left(\left( \begin{array}{c} x_1\\ \vdots \\ x_n \end{array} \right) \right) = \left( \begin{array}{cccc} f_{1\sigma(1)}(x_{\sigma(1)})\\ \vdots \\ f_{n\sigma(n)}(x_{\sigma(n)} ) \end{array} \right) $$
$$ f^{-1} \left(\left( \begin{array}{c} y_1\\ \vdots \\ y_n \end{array} \right) \right) = \left( \begin{array}{c} f_{\sigma^{-1} (1) 1 }^{-1}(y_{\sigma^{-1} (1)})\\ \vdots \\ f_{\sigma^{-1} (n) n }^{-1}(y_{\sigma^{-1} (n)}) \end{array} \right) $$
One of the cases considered in the opening question corrosponds to the case $n=2$, $\sigma(i)=i$.
Explicit example:
Consider the case $n=m=2$ where we have a 'matrix'
$$ \left( \begin{array}{cc} f_{11}(\star) & f_{12}(\star) \\ f_{21}(\star) & f_{22}(\star) \end{array} \right) = \left( \begin{array}{cc} 0 & \sin(\star) \\ \cos(\star) & 0 \end{array} \right) $$ so that there is exactly one non-zero element in each row and column. The action of this matrix can be summed up in a function $$ f \left(\left( \begin{array}{c} x_1\\ x_2 \end{array} \right) \right) = \left( \begin{array}{cc} f_{11}(\star) & f_{12}(\star) \\ f_{21}(\star) & f_{22}(\star) \end{array} \right) \left( \begin{array}{c} x_1\\ x_2 \end{array} \right) = \left( \begin{array}{cc} 0 & \sin(\star) \\ \cos(\star) & 0 \end{array} \right) \left( \begin{array}{c} x_1\\ x_2 \end{array} \right) = \left( \begin{array}{c} \sin(x_2)\\ \cos(x_1) \end{array} \right) $$ To find an inverse function we write $$ \left( \begin{array}{c} y_1\\ y_2 \end{array} \right) = \left( \begin{array}{c} \sin(x_2)\\ \cos(x_1) \end{array} \right) $$ $$ \Rightarrow \left( \begin{array}{c} x_1\\ x_2 \end{array} \right) = \left( \begin{array}{c} \arccos(y_2)\\ \arcsin(y_1) \end{array} \right) $$ $$ \Rightarrow f^{-1} \left( \left( \begin{array}{c} y_1\\ y_2 \end{array} \right) \right) = \left( \begin{array}{c} \arccos(y_2)\\ \arcsin(y_1) \end{array} \right) = \left( \begin{array}{cc} 0 & \arccos(\star)\\ \arcsin(\star) & 0 \end{array} \right) \left( \begin{array}{c} y_1\\ y_2 \end{array} \right) $$ $$ \Rightarrow \left( \begin{array}{cc} 0 & \sin(\star) \\ \cos(\star) & 0 \end{array} \right)^{-1} = \left( \begin{array}{cc} 0 & \arccos(\star)\\ \arcsin(\star) & 0 \end{array} \right) $$ The important things to note:
Tackling manipulation
Let us suppose now that we have some 'matrices' $F(\star)$ and $G(\star)$ with corresponding functions $f$ and $g$. Moreover, we have an expression $$ y = F(\star) G(\star) x $$ where $x$ and $y$ are column vectors. If we can find inverse functions for $f$ and $g$ then we can perform manipulations $$ y = F(\star) G(\star) x = f(g(x)) $$ $$ \Rightarrow f^{-1}(y) = g(x) $$ $$ \Rightarrow g^{-1}(f^{-1}(y)) = x $$ If these inverse functions can be written in matrix form then $$ F^{-1}(\star) y = G(\star) x $$ and $$ G^{-1}(\star)F^{-1}(\star)y = x $$ Hopefully this sheds some light on your 'inverse precedence conjecture'