Let $S=\{A=\left(\begin{array}{lll}0& 1& c\\ 1& a& d\\ 1& b& e\end{array}\right):a,b,c,d,e\in <!--\; \in \; -->\{0,1\}$and $|A|\in <!--\; \in \; -->\{-<!--\; -\; -->1,1\}\},$where $|A|$denotes the determinant of 𝐴. Then the number of elements in 𝑆 is _______.

To find the number of elements in the set $S$, we need to determine the number of matrices $A$ of the form:
$A=\left(\begin{array}{ccc}0& 1& c\\ 1& a& d\\ 1& b& e\end{array}\right)$
such that $a,b,c,d,e\in \{0,1\}$ and the determinant of $A$, denoted as $\left|A\right|$, belongs to the set $\{-1,1\}$.

First, let's calculate the determinant of $A$ by expanding along the first row:
$\left|A\right|=0·(ae-bd)-1·(1·e-1·d)+c·(1·b-1·a)$
Simplifying the expression, we get:
$\left|A\right|=-(e-d)+c(b-a)=d-e+c(b-a)$

We are given that $\left|A\right|\in \{-1,1\}$. Let us analyze the possible values of $\left|A\right|$ by partitioning the cases based on the value of $c\in \{0,1\}$.

Case 1: $c=0$
If $c=0$, then the determinant becomes:
$\left|A\right|=d-e$
For $\left|A\right|\in \{-1,1\}$:
- If $d-e=1$, since $d,e\in \{0,1\}$, we must have $d=1$ and $e=0$ (1 choice).
- If $d-e=-1$, we must have $d=0$ and $e=1$ (1 choice).
Thus, there are $2$ valid pairs for $(d,e)$.
Since the values of $a$ and $b$ do not affect the determinant in this case, $a$ and $b$ can be chosen independently from $\{0,1\}$.
This gives $2^2=4$ choices for $(a,b)$.
Therefore, the number of matrices in Case 1 is:
$N_1=2·4=8$

Case 2: $c=1$
If $c=1$, then the determinant becomes:
$\left|A\right|=d-e+b-a$
Let us define $x=d-e$ and $y=b-a$. The values that $x$ and $y$ can take depend on the pairs $(d,e)$ and $(b,a)$:
- Value is $1$: for pairs $(1,0)$ (1 way).
- Value is $0$: for pairs $(0,0)$ and $(1,1)$ (2 ways).
- Value is $-1$: for pairs $(0,1)$ (1 way).
We want to find the number of solutions to $x+y\in \{-1,1\}$:

Subcase 2.1: $x+y=1$
- If $x=1$ and $y=0$: there is $1$ way to choose $(d,e)$ and $2$ ways to choose $(b,a)$ ⇒ $1·2=2$ ways.
- If $x=0$ and $y=1$: there are $2$ ways to choose $(d,e)$ and $1$ way to choose $(b,a)$ ⇒ $2·1=2$ ways.
Total for this subcase = $2+2=4$ ways.

Subcase 2.2: $x+y=-1$
- If $x=-1$ and $y=0$: there is $1$ way to choose $(d,e)$ and $2$ ways to choose $(b,a)$ ⇒ $1·2=2$ ways.
- If $x=0$ and $y=-1$: there are $2$ ways to choose $(d,e)$ and $1$ way to choose $(b,a)$ ⇒ $2·1=2$ ways.
Total for this subcase = $2+2=4$ ways.

Therefore, the total number of matrices in Case 2 is:
$N_2=4+4=8$

Total elements in set $S$:
Summing the possibilities from both cases:
$N\left(S\right)=N_1+N_2=8+8=16$

Thus, the number of elements in the set $S$ is $16$.