It is a well-know observation that, given two points $x_1,x_2 \in \mathbb{R}^d$, the path signature associated to their linear interpolation is given by the tensor exponential. Precisely, if $\Delta x$ denotes the linear interpolation $x_1 \rightarrow x_2$, then $$S(\Delta x) = \left(1 , \Delta x, \frac{\Delta x^{\otimes 2}}{2!}, \frac{\Delta x^{\otimes 3}}{3!}, ... \right) =: \exp_\otimes(\Delta x) \in T((\mathbb{R}^d)),$$ where $S(\cdot)$ denotes the signature map and $T((\mathbb{R}^d))$ is the extended tensor algebra. Actually, the signature takes value in a subset $T^1$ of $T((\mathbb{R}^d))$ which can be endowed with an inner product, and thus becomes a Hilbert space if one takes the appropriate completions. I refer to this work for a proper definition, but I assume most readers interested in this post are familiar with this.
Now, instead of considering just a line segment, let us consider a piecewise linear path. Specifically, given a finite number of points $x_1,...,x_n$ in $\mathbb{R}^d$, we consider the path corresponding to successive linear interpolations, i.e. $x_1 \rightarrow x_2$ concatenated with $x_2 \rightarrow x_3$, which, in its turn, we concatenate with $x_3 \rightarrow x_4$, and so on. Let us denote this piecewise linear path by $\textbf{x}$.
Thanks to Chen's identity, we know that the signature of $\textbf{x}$ is given by multiplying in the extended tensor algebra the signatures associated to each interpolation, i.e. $$S(\textbf{x}) = \exp_\otimes(\Delta_1 x) \ \otimes \ \exp_\otimes(\Delta_2 x) \ \otimes \ ... \ \otimes \exp_\otimes(\Delta_{n-1} x),$$ where $\Delta_i x$ denotes the interpolated path $x_i \rightarrow x_{i+1}$, and $\otimes$ is the multiplication defined in $T((\mathbb{R}^d))$.
Lastly, let $\textbf{y}$ denote some other piecewise linear interpolation of $y_1,...,y_n\in \mathbb{R}^d$, and let $\langle \cdot, \cdot \rangle_{T^1}$ denote the aforementioned inner product. My question is:
Can we express $\langle S(\textbf{x}), S(\textbf{y}) \rangle_{T^1}$ via the simpler inner products $\langle \exp_\otimes(\Delta_i x), \exp_\otimes(\Delta_j y) \rangle_{T^1}$ with $i,j \in \{1,...,n-1\}$?
Firstly, you need to define an inner product in $T^1(V)$. As far as I am aware this is defined by summing all of the component-wise inner products for each tensor power; but, this is only guaranteed to converge in the range of the signature thanks to the factorial decay. Ilya's work that you link doesn't define the inner product at all (ctrl+F for "inner, dot, scalar" yielded nothing). I will assume this definition I have just mentioned since we are working with signatures.
To get a positive answer to your question, something that would help is if the algebra product $\otimes$ interacts well with the scalar product, in the sense $<a\otimes b, c\otimes d> = <a, c> <b, d>$
Unfortunately, this doesn't even work at the first level.
$<a\otimes b, c\otimes d> = <(1,a^1+b^1, ...), (1,c^1+d^1, ...)>$ $<(1,\sum_{i=0}^d a_i^1+b_i^1, ...), (1,\sum_{i=0}^d c_i^1+d_i^1, ...)>$ $=<(1,\sum_{i=0}^d a_i^1+b_i^1, ...), (1,\sum_{i=0}^d c_i^1+d_i^1, ...)>$ $=1+\sum_{i=0}^d (a_i^1+b_i^1)(c_i^1+d_i^1)+...$
whereas
$<a,c> <b , d> = (1+\sum_{i=0}^d a_i^1 c_i^1 +...)(1+\sum_{i=0}^d b_i^1 d_i^1 +...)$
so there's no way of "retrieving" the term $\sum_{i=0}^d a_i^1 d_i^1$ from the LHS in the RHS. This is just illustrative.
Now, to answer your question succinctly: in fact, you cannot retrieve $\sum_{i=0}^d a_i^1 d_i^1$ at all from $<a, c>$ and $<b, d>$ (can you see why?). As a result: no, you cannot retrieve the whole inner product of the algebra product $\otimes$ of the pieces (of your path) only from the the inner products of the pieces.