First solution (Nikolai Beluhov) Let $k = a_0 = b_0 = c_0 = d_0$. We let $f, g, h$ be great functions for $(A, B), (B, C), (C, D)$ and write the following infinite array:
$$\left[ \begin{array}{ccccccccc} \vdots & \vdots & b_3 & \vdots & \vdots & & & \\ \cdots & g(2,2) & g(2,1) & b_2 & f(1,2) & f(2,2) & \cdots & \\ \cdots & g(1,2) & g(1,1) & b_1 & f(1,1) & f(2,1) & \cdots & \\ c_3 & c_2 & c_1 & k & a_1 & a_2 & a_3 & \\ \cdots & h(2,1) & h(1,1) & d_1 & & & & \\ \cdots & h(2,2) & h(1,2) & d_2 & & & & \\ \vdots & \vdots & d_3 & & & & \ddots & \end{array} \right]$$
The greatness condition is then equivalent to saying that any $2 \times 2$ sub-grid has determinant $\pm 1$ (the sign is $+1$ in two quadrants and $-1$ in the other two), and we wish to fill in the lower-right quadrant. To this end, it suffices to prove the following.
Lemma
Suppose we have a $3 \times 3$ sub-grid
$$\begin{bmatrix} a & b & c \\ x & y & z \\ p & q & \end{bmatrix}$$
satisfying the determinant conditions. Then we can fill in the ninth entry in the lower right with an integer while retaining greatness.
Proof. We consider only the case where the $3 \times 3$ is completely contained inside the bottom-right quadrant, since the other cases are exactly the same (or even by flipping the signs of the top row or left column appropriately).
If $y = 0$ we have $-1 = bz = bx = xq$, hence $qz = -1$, and we can fill in the entry arbitrarily.
Otherwise, we have $bx \equiv xq \equiv bz \equiv -1 \pmod y$. This is enough to imply $qz \equiv -1 \pmod y$, and so we can fill in the integer $\frac{qz+1}{y}$. $\square$

Second solution (Ankan Bhattacharya) We will give an explicit classification of great sequences:
Lemma
The pair $(A, B)$ is great if and only if $a_0 = b_0$, $a_0 \mid a_1b_1 + 1$, and $a_n \mid a_{n-1} + a_{n+1}$ and $b_n \mid b_{n-1} + b_{n+1}$ for all $n$.
Proof of necessity. It is clear that $a_0 = b_0$. Then $a_0f(1,1) - a_1b_1 = 1$, i.e. $a_0 \mid a_1b_1 + 1$. Now, focus on six entries $f(x,y)$ with $x \in \{n-1, n, n+1\}$ and $y \in \{0, 1\}$. Let $f(n-1,1) = u$, $f(n,1) = v$, and $f(n+1,1) = w$, so
$$\begin{aligned} va_{n-1} - u a_n &= 1, \\ wa_n - v a_{n+1} &= 1. \end{aligned}$$
Then
$$u + w = \frac{v(a_{n-1} + a_{n+1})}{a_n}$$
and from above $\gcd(v, a_n) = 1$, so $a_n \mid a_{n-1} + a_{n+1}$; similarly for $b_n$. (If $a_n = 0$, we have $v a_{n-1} = 1$ and $v a_{n+1} = -1$, so this is OK.) $\square$
Proof of sufficiency. Now consider two sequences $a_0, a_1, \dots$ and $b_0, b_1, \dots$ satisfying our criteria. We build a great function $f$ by induction on $(x, y)$. More strongly, we will assume as part of the inductive hypothesis that any two adjacent entries of $f$ are relatively prime and that for any three consecutive entries horizontally or vertically, the middle one divides the sum of the other two.
First we set $f(1,1)$ so that $a_0 f(1,1) = a_1 b_1 + 1$, which is possible.
Consider an uninitialized $f(s,t)$; without loss of generality suppose $s \ge 2$. Then we know five values of $f$ and wish to set a sixth one $z$, as in the matrix below:
$$\begin{pmatrix} u & x \\ v & y \\ w & z \end{pmatrix}$$
(We imagine $a$-indices to increase southwards and $b$-indices to increase eastwards.) If $v \ne 0$, then the choice $y \cdot \frac{u+w}{v} - x$ works as $u y - v x = 1$. If $v = 0$, it easily follows that $\{u, w\} = \{1, -1\}$ and $y = w$ as $y w = 1$. Then we set the uninitialized entry to anything.
Now we verify that this is compatible with the inductive hypothesis. From the determinant $1$ condition, it easily follows that $\gcd(w, z) = \gcd(v, z) = 1$. The proof that $y \mid x + z$ is almost identical to a step performed in the “necessary” part of the lemma and we do not repeat it here. By induction, a desired great function $f$ exists. $\square$
We complete the solution. Let $A, B, C$, and $D$ be integer sequences for which $(A, B)$, $(B, C)$, and $(C, D)$ are great. Then $a_0 = b_0 = c_0 = d_0$, and each term in each sequence (after the zeroth term) divides the sum of its neighbors. Since $a_0$ divides all three of $a_1b_1 + 1$, $b_1c_1 + 1$, and $c_1d_1 + 1$, it follows $a_0$ divides $d_1a_1 + 1$, and thus $(D, A)$ is great as desired.