Since $\alpha,\boldsymbol{e}$ are both constant, let $c := \alpha\boldsymbol{e}$.
Let $J_i$ be the i'th row of the matrix $J$, i.e. $J:=\pmatrix{J_1\\\vdots\\ J_n}$.
Then from the definition of the matrix multiplication follows that
$$
Ju = \pmatrix{J_1 u \\\vdots\\ J_n u}
$$
Therefore
$$
\min_{u} | Ju+c|_1
=
\min_{u} \sum_{i=1}^n |J_i u+c_i|
$$
Now, we can use your linked question:
It tells us that we take every summand $|J_i u+c_i|$, replace it with a new variable $x_i$ in the objective function and add the two constraints $x_i\ge (J_i u+c_i)$ and $x_i \ge -(J_i u+c_i)$.
Now our minimization problem is
$$
\min_{u,x_1,...,x_n} \sum_{i=1}^n x_i
$$
such that
$$\forall i\in \{1,...,n\}:\quad x_i\ge (J_i u+c_i) ,x_i \ge -(J_i u+c_i)$$
And from now it's reverse fitting:
Defining $x:= \pmatrix{x_1\\\vdots\\x_n}$ let's us reformulate to:
$$
\min_{u,x} \mathbb{1}^T x
$$
such that
$$x\ge J u+c ,x \ge -J u-c$$
Now we make the substitution $ x\leftarrow 2y-Ju-c$ and obtain:
$$
\min_{u,y} \mathbb{1}^T (2y-(Ju+c))
$$
such that
$$y\ge Ju+c ,y \ge 0$$
, which is the claimed formula.
Regarding the comment:
Substitutions like this are something general to equation systems. There are lots of small tricks involved, depending on what you want to do; Unluckily, I know of no source with gathers them.
I'll try for a somewhat general explanation:
First, rewrite the optimization problem using set notation:
$$
\min\left\{ 1^T x \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land x\ge Ju+c \land x \ge -Ju-c \right\}
$$
In other words, we have a function selecting on a set, and the function selects a specific element of this set. Let's call the set $M_0$.
Now, a substitution is an action altering the set. In general, this happens by adding, removing and/or altering the constraints on the right side of the set (i.e. after the $|$). You can do anything here (but if you add variables, then you should also add a constraint stating which set they do come from), but for your action to be of use for you, you want in general to be able to prove afterwards, that your resulting set, let's call it $M_1$, has one of the following three relations to $M_0$:
a) $M_0\subseteq M_1$
b) $M_0=M_1$
c) $M_0\supseteq M_1$
Then a) implies $\min M_1 \le \min M_0$,
b) implies $\min M_1 = \min M_0$,
and c) implies $\min M_1 \le \min M_0$.
Armed with this knowledge, we can know apply this to our substitution: We do this by adding the equation $x=2y-Ju-c$ to $\left\{ 1^T x \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land x\ge Ju+c \land x \ge -Ju-c \right\}$, obtaining
$$
\min\left\{ 1^T x \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land x\ge Ju+c \land x \ge -Ju-c \land x=2y-Ju-c\right\}
$$
Now linear algebra tells you that for every $x$, there is a unique solution of $x=2y-Ju-c$ for $y$. Therefore we have
$$
\left\{ 1^T x \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land x\ge Ju+c \land x \ge -Ju-c \right\}\\=\\
\left\{ 1^T x \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land x\ge Ju+c \land x \ge -Ju-c \land x=2y-Ju-c\right\}
$$,
because any $x$ which satisfied the set restriction of the LHS still does in the RHS if we choose $y = (x-Ju-c)/2$, and in turn for a pair $(x,y)$ to satisfy the set restriction of the RHS, it has to hold that $y = (x-Ju-c)/2$.
When working with sets, any equation in the set restriction can be substituted into any other equation in the set restriction or the LHS of the set.
Doing that, we obtain:
$$
\left\{ 1^T x \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land x\ge Ju+c \land x \ge -Ju-c \land x=2y-Ju-c\right\}
\\=\\
\left\{ 1^T (2y-Ju-c) \mid u\in\mathbb{R}^n \land x\in \mathbb{R}^m\land 2y-Ju-c\ge Ju+c \land 2y-Ju-c \ge -Ju-c \land x=2y-Ju-c\right\}
$$
, and all that's left to do is simplify each part on its own.
