Look how the method works for the number $abcde$. You substract $e$ from $abcd$, getting a'b'c'd', and you apply the method to a'b'c'd'. If the method works for a'b'c'd', it yields n' such that a'b'c'd'=11\times n'. But a'b'c'd'0 is a'b'c'd'\times 10=11\times 10\times n' and abcde=a'b'c'd'\times 10+e\times 11 hence abcde=11\times 10\times n'+11\times e is 11\times(10\times n'+e). This proves that n=10\times n'+e is indeed the correct answer for $abcde$.
Edit (Upon OP's request, the same proof, with more apparatus but with zero more mathematics.)
Look how the method works for the number $N=a_ka_{k-1}\cdots a_2a_1a_0$ with $k\geqslant1$. You substract $a_0$ from $a_ka_{k-1}\cdots a_2a_1$, getting $M=b_kb_{k-1}\cdots b_2b_1$, and you apply the method to $M$. If the method works for $M$, it yields $m$ such that $M=11\times m$. But $b_kb_{k-1}\cdots b_2b_10$ is $M\times 10=11\times 10\times m$ and $N=M\times 10+a_0\times 11$ hence $N=11\times 10\times m+11\times a_0$ is $11\times(10\times m+a_0)$. This proves that $n=10\times m+a_0$ is indeed the correct answer for $N$ if $m$ was the correct answer for $M$. A recursion on the number of digits of $N$ yields the result.