Why can we modular reduce arguments of sums and products in modular arithmetic?

The Congruence Sum & Product Rules imply that congruences are preserved if we replace $\rm\color{#c00}{arguments}$ of sums and products by $\rm\color{#0a0}{congruent}$ arguments. Applying this inductively shows the same holds true for arbitrary expressions composed of sums & products, i.e. polynomial expressions, yielding the linked (univariate) Polynomial Congruence Rule.

Yours is the special case below for the polynomial $\,10a+b,\,$ for modulus $\, n = 11,\,$ and $\,x' := x\bmod n = x\%11$.

$\left.\begin{align}{\bf Theorem}\ \ \bmod n\!:\,\ \color{#c00}{a'}\equiv \color{#0a0}a\\ b'\equiv b\end{align}\right\}\, $ $\Rightarrow$ $\,\ \begin{align} &10\,\color{#c00}{a'}+b'\\ \equiv\ &10\,\color{#0a0}a\,+\,b\end{align}$

$\begin{align}{\bf Proof}\qquad a'&\equiv a\qquad\quad\ \, \text{by hypothesis}\\ 10a'&\equiv 10a\qquad\ \ \text{by the Congruence Product Rule}\\ b'&\equiv b\qquad\quad\ \ \text{by hypothesis}\\ \Rightarrow\ 10a'+b'&\equiv 10a+b\ \ \ \text{by the Congruence Sum Rule} \end{align}$

Remark $ $ To get the exact form of your result apply a final $\bmod 11\,$ to the above to convert it from a congruence relation to a mod operation (remainder), using the following $$ a\equiv b\!\!\!\pmod{n}\iff (a\bmod n) = (b\bmod n) $$

Generally this is the easiest way to prove identities about mod operations, i.e. use more flexible congruences to first prove the analogous congruence relation, then apply a final mod operation to get (canonical / normal) remainders (or residues).