There exists a minimal subshift $X$ with a point $x \in X$ such that $x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$?

We can produce such a subshift by a standard hierarchical construction. Let $w_{0,0} = 01$ and $w_{0,1} = 011$. For each $k \geq 0$, define $w_{k+1,0} = w_{k,0} w_{k,0} w_{k,1}$ and $w_{k+1,1} = w_{k,0} w_{k,1} w_{k,1}$. Define $X$ by forbidding each word that doesn't occur in any $w_{k, b}$. Since each $w_{k+1,b}$ contains both $w_{k,0}$ and $w_{k,1}$, it's easy to show that $X$ is minimal.

For $b \in \{0,1\}$, define $x^b \in X$ as the "limit" of $(w_{k,b})_{k \geq 0}$ such that the central $01$ or $011$ is at the origin. Because of the way we defined $w_{k,0}$ and $w_{k,1}$, the only difference between the words, and hence the limiting configurations, is that single extra $1$. Concretely, $x^0$ and $x^1$ look like $$ \cdots 0101011\;0101011\;01011011\;0101011\;010.1011\;01011011\;0101011\;01011011\;01011011 \cdots $$ and $$ \cdots 0101011\;0101011\;01011011\;0101011\;010.11011\;01011011\;0101011\;01011011\;01011011 \cdots $$ (Spaces added for clarity.)

It is well-known that the Chacon substitution $\tau$ defined by $\tau(0) = 0010$, $\tau(1) = 1$ produces a minimal subshift, when you take the legal words to be the words that appear in some $\tau^n(0)$. The two-sided fixed point from $0.0$ is $x.y = {...0010001010010.0010001010010...}$ and the one from $0. 10$ is $x.1y$. So they are both in $X$.

Now apply the additional substitution $\alpha(0) = 01$, $\alpha(1) = 1$ and you still get a minimal subshift $Y$ as image (this is a flow equivalence onto its image). We have $\alpha(x).\alpha(y) \in Y$ and $\alpha(x) . 1 \alpha(y) \in Y$, and since $\alpha(x)$ ends with $1$ this is a pair of the kind you are asking for.