How to recover a message with shared modulus for textbook RSA
what I am doing wrong?

Accepting as fact a recipe with an equation, rather than deriving it.
Illustration: « Then we obtain $m$ as ${c_1}^a + {c_2}^b \bmod n$ » is stated rather than derived. And wrong.
As an aside the question reverses $a$ and $b$ (or is it $c_1$ and $c_2$, or $e_1$ and $e_2$): they are correct per the official solution which asks $a\,e_1 + b\,e_2 = 1$, but the calculation then made assumes and states $a\,e_2 + b\,e_1 = 1$.
Hint to derive the correct equation:

Prove that $m^1\bmod n=m$, by combining the definition¹ of raising to an integer exponent, the assumed range of $m$ in textbook RSA, and the definition² of the $\bmod$ operator.
In this equation, substitute $1$ with $a\,e_1 + b\,e_2$ where $a$ and $b$ are the Bezout coefficients such that $a\,e_1 + b\,e_2=1$.
Properly use properties³ of modular exponentiation. In cryptanalysis it's OK to assume plausible preconditions when that's necessary, but it remains good to verify them (or the result derived) in the end.


Suggestion: apply this technique with a slightly larger $n=14835196795348830319$, $(e_1,e_2)=(3,5)$, and $(c_1,c_2)=(14562201346830272020,1832973312396331965)$. As a bonus find the menu by expressing $m$ in hex. The point of these larger numbers is that guessing the Bezout coefficients and inverse is harder, and slightly more computer skills are needed.
More mathematically interesting, for a different method is needed, which would still work if $n$ was too large to factor directly from its value: with the same $(n,e_1,e_2)$ find $m$ for $(c_1,c_2)=(11810011337245959646,2207245693327700143)$.

Note: in actual use, RSA encryption

Does not reuse the same $n$ among several public keys, making this particular attack fail.
Does not directly encode the message as $m$, but rather adds randomness to the message in order to form $m$. That makes it extremely unlikely that $m$ is reused (it would be enough to allow some other attacks including with different $n$).
Uses $n$ with several hundred decimal digits in order to resist factorization. $n=143$ can be factored mentally, and the larger $n$ is factored in a small fraction of a second using a computer, which allows to find $m$ by using the normal RSA decryption equation, without needing $c_2$ or $e_2$.


Appreciation: taking at face value the statement that the extended Euclidean algorithm was used, the question shows fair skills to apply equations and algorithms, despite the inversion of $a$ and $b$. Be assured that the goal of such exercises is not learning the equations used. It's to learn to derive the necessary equations. That's easier when one gets the hang of it, and more fun. It's more reliable, and thus gets higher marks (which is a valid sub-goal) even when the reasoning is not considered in the notation. As a skill it's more useful, more adaptable, less likely to go obsolete, or unused and forgotten.

¹ ² ³ : See this definition of $x^u\bmod k$ and related typographical conventions, or refer to a previous version of the present answer.