A well-known property of delta function is the following: $$\int\limits_{a-\epsilon}^{a+\epsilon}f(x)\delta(x-a)dx=f(a)$$ for $\epsilon > 0$.
For the case of a shifted function I would like to know if the following holds always: $$\int\limits_{a-\epsilon}^{a+\epsilon}f(x-b)\delta(x-a)dx=f(a-b)$$ for $\epsilon > 0$, $b\in \mathbb{R}$.
In general, I am wondering about the following: $$\int\limits_{a-\epsilon}^{a+\epsilon}f(g(x))\delta(x-a)dx=f(g(a))$$ for $\epsilon > 0$, $b\in \mathbb{R}$ and $g$ known real function.
Just define $h(x):= f(x-b)$ and $k(x):=f(g(x))$. Then you can use the well known property of the dirac delta, which you already stated, for the functions $h$ and $k$ and conclude the two identities:
$$ \int_{a-\epsilon}^{a+\epsilon}f(g(x))\delta(x-a)dx = \int_{a-\epsilon}^{a+\epsilon}k(x)\delta(x-a)dx = k(a) = f(g(a)) $$