Given that the cardinality of a countable set $A$ equal $\mathfrak a$, and $A$ be partitioned into singleton with elements within. Can we define the cardinality of disjoint double union with index $i\in I$ and $j\in J$, $I, J\subset \Bbb N$ such that:
$$A=\bigcup_{i=1}^{|I|}\bigcup_{j=1}^{|J|}\{a_{ij}\}$$
$$\left|\bigcup_{i=1}^{|I|}\bigcup_{j=1}^{|J|}\{a_{ij}\} \right|=\sum_{i=1}^{|I|}\sum_{j=1}^{|J|}|\{a_{ij}\}|=\sum_{i=1}^{|I|}\sum_{j=1}^{|J|}1 $$
$$=|I||J|=\mathfrak a\text{ is true?}$$
And if so, can it serve as an alternate definition of same cardinality between sets as existing a bijection?
EDIT:
The answer provided by @Hamdiken was satisfactory. Just to add the context of my motivation:
I thought of a prove that $|\{2k:k\in\Bbb N\}|=|\{2k-1:k\in\Bbb N\}|=|\Bbb N|$ like:
$$|\{2k:k\in\Bbb N\}|=\left|\lim_{n\to\infty}\bigcup_{k=1}^n\{2k\} \right| $$
$$=\lim_{n\to\infty}\left|\bigcup_{k=1}^n\{2k\} \right|=\lim_{n\to\infty}\sum_{k=1}^n|\{2k\}| $$
$$\lim_{n\to\infty}\sum_{k=1}^n 1=\lim_{n\to\infty}n=|\Bbb N| $$
The same can be done for $\{2k-1:k\in\Bbb N\}$. So given that for that case works, then I tried to prove with the same method that $|\Bbb Z|=|\Bbb N|$, so i tried to define $\Bbb Z=\{k:k\in \Bbb N \}\cup\{-k:k\in \Bbb N\}\cup\{0\}$, so:
$$|\Bbb Z|=\lim_{n\to\infty}\left|\{0\}\cup\left(\bigcup_{i=1}^2\bigcup_{k=1}^n\{(-1)^ik\}\right) \right| $$
$$=\lim_{n\to\infty} 1+\left|\bigcup_{i=1}^2\bigcup_{k=1}^n\{(-1)^ik\} \right|=1+\lim_{n\to\infty}\sum_{i=1}^2\sum_{k=1}^n|\{(-1)^ik\}| $$
$$=1+\lim_{n\to\infty}\sum_{i=1}^2\sum_{k=1}^n 1=1+\lim_{n\to\infty}2n $$
$$1+2|\Bbb N| $$
and by the info in this page, $1+2|\Bbb N|=|\Bbb N| $, and then this method would also work for $|\Bbb Q|=|\Bbb N|$. Sadly, doesn't for for all cardinality so I don't think it serve as a proof.
At first glance, when you develop your hypothesis, it looks like this : $$A=\bigcup_{i=1}^{|I|}\bigcup_{j=1}^{|J|}\{a_{ij}\}$$ $$=\bigcup_{i=1}^{|I|}\left(\left\{a_{i,1},a_{i,2},...,a_{i,|J|}\right\}\right)$$ $$=\left\{a_{1,1},a_{1,2},...,a_{k,1},a_{k,2},...a_{|I|,|J|}\right\}$$ This set (Given that your set $A$ is of finite cardinality $\mathfrak a$) can be rearranged into a matrix $M\in\mathcal M_{|I|,|J|}(\mathbb K)$ such that
$$M= \pmatrix{a_{1,1}&a_{1,2}&...&a_{1,|J|}\\ a_{2,1}&a_{2,2}&...&a_{2,|J|}\\ \vdots&\vdots&\ddots&\vdots\\ a_{|I|,1}&a_{|I|,2}&\cdots&a_{|I|,|J|} } $$ The question is : Is it always possible to arrange $\mathfrak a$ elements into a matrix ? (Consider the case $\mathfrak a =17$).
Hope this answers your question.