Is it possible to extract any encoded $x, y \in \mathbb{N^*}$ from $z=ax + by$ Is there any specific $a, b \in \mathbb{R}$, $\forall x,y \in
\mathbb{N^*}$, take $z=a\cdot{}x+b\cdot{}y$ (then $z\in\mathbb{R}$),
we can always extract $a,b$ from $z$.

Here below are some trials I made but they don't meet the all the requirements.
Method 1:
Take $a=10^k,b=1,k\in\mathbb{N^*}$. Then we can encode any $0\leq
x,y\lt 10^k$. e.g. $42\cdot10^2+32=4232$, in which it's possible to
extract $42$ and $32$ from the result $4232$.
The limitation is that in this way only a limited number of $x,y$ are allowed.
Method 2:
Take $$a=1,b=\lim_{k\to{}\infty}10^{-k}$$, any $x,y\in \mathbb{N^*}$ are
allowed now e.g. $42\cdot
1+32\cdot0.\underbrace{000\dots000}_\text{infinite
zeros}1=42.\underbrace{000\dots000}_\text{infinite zeros}32$
However, the problem is that $b$ isn't valid in $\mathbb{R}$.
Method 3:
Take $a=1, b=\mathrm{i}$. The result will be a complex number. But
this doesn't meet the condition $a,b \in \mathbb{R}$.
Ideas:
I also considered to use $\pi$ or $\mathrm{e}$ or other irrational
numbers like $\sqrt{2}$ to be $a$ and $b$. But I don't know if it is
flexible to extract the result in some methods.

This question rose from a trial to implement cross product of two
vectors directly on a scientific calculator. A cross product of
$\vec{a}$ and $\vec{b}$ can be calculated by:
$$\vec{a}\times\vec{b}= \det\left(
\begin{array}{ccc}
\vec{\mathrm{i}} & \vec{\mathrm{j}} & \vec{\mathrm{k}} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array} \right) =
m_1\cdot\vec{\mathrm{i}}+m_2\cdot\vec{\mathrm{j}}+m_3\cdot\vec{\mathrm{k}}=
\vec{z}$$, hence we can have $m_1,m_2,m_3\in\mathbb{Q}$ if
$a_1,a_2,a_3,b_1,b_2,b_3\in\mathbb{Q}$.
However the calculator can only accept numeric values in a matrix.
Therefore I cannot use any symbols in it. I conceived that whether
there I can take special values in $\mathbb{R}$ instead of
$\vec{\mathrm{i}},\vec{\mathrm{j}},\vec{\mathrm{k}}$, so in the
resulting value I can recover $m_1$,$m_2$ and $m_3$ from it. This is a
interest question to through inquiry.
To make the conditions simpler, I limited the range of $m_i$ from
$\mathbb{Q}$ into $\mathbb{N^*}$, and reduced the number of terms from
$3$ to only $2$. The proposition is now weaker but still equally
interesting.

The uniqueness you want is guaranteed by linear independence over $\mathbb Q$. 
If you choose $x$ rational and $y$ irrational, then you are good. If $ax+by=0$ for rational $a,b$, then $by=-ax$; but the left hand side is irrational, while the right hand side is rational. Unless $b=0$, and then $a=0$. 
And you can do the same with 3 numbers, by taking $x$ rational, $y$ irrational but algebraic, and $z$ trascendental. For instance, $x=1$, $y=\sqrt2$, $z=\pi$. Then, if $a+b\sqrt2+c\pi=0$, we have $c\pi=-a-b\sqrt2$, the left-hand-side trascendental and the right-hand-side algebraic, so $c=0$. Now we have $a+b\sqrt2=0$ and we can reason as in the case of two numbers. 
Now, regarding your goals, you cannot use irrational numbers in a calculator. And a second problem is that it could be hard to find $a,b,c$ given $a+b\sqrt2+c\pi$. 
