Assume that the first claim is false, we then prove the truth of the second claim. Letting $n = p_1^{\alpha_1} \dots p_t^{\alpha_t}$ it follows that $f(n) = f(p_1) \dots f(p_t)$. Thus, the value of $f$ is only depending upon the values of $f$ at prime points. Hence, it suffices to prove that for all large enough $p$, $f(p) = 1$. Let us define the set $A$ as follows:
$$A = \{an + b : f(an + b) = 1, n \in \mathbb{N}\}$$
The first criterion ensures that $A$ is nonempty. We shall prove that every large enough prime has a multiple in that set. Take $n_0$, $m_0$ such that $an_0 + 1 = (ak + b)\varphi(a)$ and $cm_0 + 1 = (ak + b)\varphi(c)$ then $f(an_0 + 1) = g(cm_0 + 1) = 1$.

Then, if $an + b \in A$ then
$$1 = f((an + b)(an_0 + 1)) = f(a(ann_0 + bn_0 + n) + b) = g(c(ann_0 + bn_0 + n) + d).$$
Hence,
$$\begin{align*} 1 &= g(c(ann_0 + n + bn_0) + d)g(cm_0 + 1) \\ &= g(c(cm_0(ann_0 + n + bn_0) + (ann_0 + n + bn_0) + dm_0) + d) \\ &= f(a((cm_0 + 1)(ann_0 + n + bn_0) + dm_0) + b). \end{align*}$$
That is, $(cm_0 + 1)(an_0 + 1)(an + b) + (ad - bc)m_0 = s(an + b) + t \in A$. That is, for all $x \in A$ we shall have $sx + t \in A$. By choosing another $n_0$ we can find another $s_1$ such that $s_1x + t \in A$ for all $x \in A$. Now, take a large prime $p$ and let us denote by $b_1, \dots, b_k$ the residues of elements of $A$ modulo $p$. Hence,
$$\{s_0 b_1 + t, \dots, s_0 b_k + t\} \equiv \{s_1 b_1 + t, \dots, s_1 b_k + t\} \equiv \{b_1, \dots, b_k\} \pmod{p}.$$
That is, $s_0(\sum_{i=1}^k b_i) + tk \equiv s_1(\sum_{i=1}^k b_i) + tk \equiv \sum_{i=1}^k b_i \pmod{p}$. Therefore, $\sum_{i=1}^k b_i \equiv 0 \pmod{p}$ and $k = p$. That is, $0 \in A$. It yields that, there is a number of the form $an + b$ in $A$ that is a multiple of $p$. So, $f(an + b) = 1$ implies that $f(p) = 1$. By the same argument we find that $g(p) = 1$. Taking $D = \prod_{q < p} q$, we are done. $\blacksquare$

Comment 1. The criterion that $a$, $b$, $c$, $d$ are pair-wise coprime can be reduced to the condition $ad \neq bc$.

Comment 2. We can indeed prove that there is a positive integer $D$ such that for all positive integers $n$, $m$, $\text{gcd}(nm, D) = 1$ we have $f(n) = g(m) = 1$. Here we provide the sketch of the solution. If $f(an + b) = g(cn + d)$ for some positive integer $n$. Assume that $\text{gcd}(a, b) = \text{gcd}(c, d) = 1$. Let $p$ be a prime dividing $an + b$ then $\text{gcd}(a, p) = 1$ hence $P_t = p^{t\varphi(a)} = aT_t + 1$ and $f(P_t) = 1$. Hence,
$$g(cn + d) = f(an + b) = f(P_t)f(an + b) = f(a(nP_t + bT_t) + b) \quad (1)$$
$$= g(c(nP_t + bT_t) + d), \quad (2)$$
for all positive integers $n$, $t$. Let $S_f = \{n : f(n) = 1\}$, $S_g = \{n : g(n) = 1\}$.
It follows that $c(nP_t + bT_t) + d \in S_g$ for each $t$. Now, we can prove the following interesting lemma that can be proved by use of techniques presented above.

Lemma. Let $a$, $b$, $c$, $d$ be integers such that $a$, $c > 0$, $ad \neq bc$. We define the set $S$ as follows:
i. If $x$, $y \in S$, $\text{gcd}(x, y) = 1$, then $xy \in S$,
ii. If $x \in S$, $y \mid x$ and $\text{gcd}(\frac{x}{y}, y) = 1$, then $y \in S$,
iii. $an + b \in S$ if and only if $cn + d \in S$.

Let $r$ be a prime number that doesn't divide $ac$ and $m$ be a positive integer such that $\{r^m, r^{m+1}, \dots\}$ is a subset of $S$ and there is a positive integer $n$ such that $cn + d \in S$, for some positive integer $n$, there is a prime number $q$ that doesn't divide $ac$ and $\{1, q, q^2, \dots\} \subseteq S$ and all integers that are coprime to $D = (ad - bc)\gcd(a, c)$ are in $S$.

According to this interesting lemma, $N_t = \{n \in \mathbb{N} : \gcd(n, c(ad - bc)T_t) \subseteq S_g, \text{ for all } t \in \mathbb{N}\}$. Now, using Chinese Remainder Theorem (CRT) to choose $m_0$ such that $\gcd(am_0 + 1, (ad - bc)T_1) = 1$ and $\gcd(m_0, T_1) = 2$. Hence, $\{n \in \mathbb{N} : \gcd(n, c(ad - bc)m_0) \subseteq S_g$ and therefore, $\{n \in \mathbb{N} : \gcd(n, 2c(ad - bc)) \subseteq S_g\}$. Continuing this way, we can prove that $\{n \in \mathbb{N} : \gcd(n, 2a(ad - bc)) \subseteq S_f$. The rest is clear.