A (two-sided) topological Markov shift is a system $(W,S)$ defined by a matrix $A=(a_{i,j})_{i,j=1}^k$ with entries in $\left\{0,1\right\}$ as follows. $W$ is the subset of $\left\{1,\ldots,k\right\}^{\mathbb{Z}}$ consisting of sequences $(x_n)_{n\in\mathbb{Z}}$ with the property that $a_{x_n,x_{n+1}}=1$ for all $n$ and $S$ is simply the transformation that shifts a sequence one unit to the left. Then the topological entropy is $h(W,S)=\log\lambda$, where $\lambda$ is the largest positive eigenvalue of $A$. The same holds for one-sided Markov shift (i.e. when considering only the sequences in, for example, $\left\{1,\ldots,k\right\}^{\mathbb{N}}$).
Now to my question. Assume we consider two symmetric one-sided topological Markov shifts, one with the left-shift $S_L$, the other with the right-shift $S_R$. More precisely, let
$W_L\subset\left\{1,\ldots,k\right\}^{\mathbb{Z>0}}$ with some matrix $A_L$ such that $S_L$ acts as the left shift on $W_L$. Assume $h(W_S,S_L)=\ln\lambda$ with $\lambda$ being the largest positive eigenvalue of $A_L$.
Moreover, let $W_R\subset\left\{1,\ldots,k\right\}^{\mathbb{Z}_{\leq 0}}$ with some matrix $A_R$ such that $S_R$ acts as the right shift on $W_R$ and assume that $h(W_R,S_R)=\ln\lambda$.
Then, in case both systems do not depend on each other, we have $$ h(W_R\times W_L,S_R\times S_L)=h(W_R,S_R)+h(W_L,S_L)=2\ln\lambda. $$
Now, suppose we have some function $a(w_R,w_L)\in\left\{-1,0,1\right\}$ with $w_R\in W_R, w_L\in W_L$ and we want to compute or estimate the topological entropy $$ h(W_R\times W_L,S_R^{1-a(\cdot,\cdot)}\times S_L^{1+a(\cdot,\cdot)}). $$
(That is, if, for example, $a(w_R,w_L)=1$, we are not shifting $w_R$ at all, but shift $w_L$ twice to the left.)
Question:
Since for each choice of $w_R\in W_R$ and $w_L\in W_L$, we have $$ (1-a(w_R,w_L)) + (1+a(w_R,w_L))=2, $$ and we have, in general, that $h(S_R^0)=h(\text{id})=0, h(S_R^2)=2h(S_R)$ and, similarly, $h(S_L^0)=0$ and $h(S_L^2)=2h(S_L)$, do we have again $$ h(W_R\times W_L,S_R^{1-a(\cdot,\cdot)}\times S_L^{1+a(\cdot,\cdot)})=2\ln\lambda $$ or at least get an upper estimate $$ h(W_R\times W_L,S_R^{1-a(\cdot,\cdot)}\times S_L^{1+a(\cdot,\cdot)})\leq 2\ln\lambda $$ or lower estimate $$ h(W_R\times W_L,S_R^{1-a(\cdot,\cdot)}\times S_L^{1+a(\cdot,\cdot)})\geq 2\ln\lambda? $$
For me that would be not too absurd since for each argument $(w_R,w_L)$ we make two shifts, effectively: If $a(w_R,w_L)=0$, we do a single shift in each component and if $a(w_R,w_L)=-1$ or $a(w_R,w_L)=1$, we do a double shift in one component and no shift in the other.
Detailing my answer above:
The answer is yes, provided that the function $a$ is continuous. In fact, the question doesn't even make sense otherwise since the entropy is defined for continuous maps.
As usual, you only need to count the number of cylinder sets of some length, and in this case the number won't change if you count around zero. So indeed the number of cylinders of a given length is the same as when $a$ vanishes. This gives your intended result.