There is a proposition which says:
If $a$ and $b$ are two periodic elements of a monoid $(G,\cdot\,,1)$ such that $a \cdot b = b \cdot a$, then $a \cdot b$ is a periodic element of the monoid $(G,\cdot\,,1)$.
I tried to find a counterexample. I thought of matrices. I didn't find two $2 \times 2$ matrices so I tried with $3 \times 3$ but no result. I thought also to try with functions. I am searching two elements $a$ and $b$ of finite order such that $a \cdot b \neq b \cdot a$ and $a \cdot b$ is not of finite order. Since I didn't find a counterexample, I thought it may be true also the converse.
Any help to find two non commutative periodic elements such that their product is non-periodic? Is the converse true: $a \cdot b = b \cdot a$ and $a \cdot b$ periodic implies $a$ and $b$ are periodic?
You have phrased your question confusingly multiple times (I think there were multiple accidental word substitutions) but I think you are asking the following two questions:
Is it true that if $a, b \in G$ are two elements of a group which don't necessarily commute, and $a, b$ have finite order, then $ab$ must have finite order?
Is it true that if two elements $a, b \in G$ of a group commute and $ab$ has finite order then $a, b$ have finite order?
The answer to both questions is no. For the first question you can, in fact, find a counterexample in $2 \times 2$ matrices: take two reflections across two axes at an irrational angle apart (I mean an irrational multiple of $\pi$). These have order $2$ but their product is an irrational rotation which has infinite order. Explicitly you can take
$$a = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right], b = \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{array} \right]$$
where $\theta$ is an irrational multiple of $\pi$, e.g. $\theta = 1$. We have $a^2 = b^2 = 1$ but their product
$$ab = \left[ \begin{array}{cc} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \right]$$
is a clockwise rotation by $\theta$. By using other values of $\theta$ we can arrange for the product $ab$ to have any finite order. $a$ and $b$ generate, depending on the angle $\theta$, either a finite or an infinite dihedral group.
For the second question you can take $a$ to be any element of infinite order in any group, e.g. $1 \in \mathbb{Z}$, and $b = a^{-1}$, so that $ab = ba = e$.