Let $A=(a_0, a_1, a_2, a_3,...)$ and $B=(b_0, b_1, b_2, b_3,...)$ denote infinite integer sequences. Define addition coordinate-wise as $A+B=(a_0+b_0, a_1+b_1,a_2+b_2,a_3+b_3,...a_n+b_n,...)$ and multiplication as a discrete convolution of $A$ and $B$ i.e. $AB=C=(c_0,c_1,c_2,c_3,...)$ where $$c_i=\sum_{i=0}^{n}a_ib_{n-i}$$. Let $R$ denote this ring.
Similarly, let $A=(a_1, a_2, a_3, a_4,...)$ and $B=(b_1, b_2, b_3, b_4,...)$ be integer sequences, except this time, the starting index is $1$. Addition is defined the same as in $R$, but multiplication defined as the Dirichlet Convolution instead of the discrete convolution: $AB=C=(c_1,c_2,c_3,c_4,...)$ where $$c_i=\sum_{i|n}a_ib_{n/i}$$ This operation can be shown to be associative similar to how the discrete convolution is shown to be associative. Call this ring $S$.
I want to show that no homomorphism $\phi: R \rightarrow S$ exists. Here is my approach at showing an isomorphism doesn't exist. Let $\phi: R \rightarrow S$ be an isomorphism. Let $a=(1,1,1,1,1,....) \in R$ where $a_n=1$. Now $a^2=b=(1,2,3,4,5,6,...)$ where $b_n = n$.
My claim is that $b$ would also have to be a perfect square in $S$ in order for an isomorphism to exist. That is, there is some sequence $c$ such that $\phi(c)=b$. Also, let $\phi(c_2)=a$. Now $\phi(c)= \phi(a*a) = \phi(c_2)*\phi(c_2)$.
Let $T = (t_1, t_2, t_3, t_4, ...) \in S$. Now, $T^2 = (t_1^2, 2t_1t_2, 2t_1t_3, 2t_1t_4+t_2^2...)$. If $b$ was a perfect square in $S$, we must have $(1,2,3,4,...)=(t_1^2, 2t_1t_2, 2t_1t_3, 2t_1t_4+t_2^2...)$
Clearly, there are no integers such that $b_3 = 3 = 2t_1t_3$ since the right hand side is odd, but the left hand side is even.
Thus $b$ is not a square in $S$, contradicting that $\phi$ is an isomorphism. Therefore, no isomorphism between $R$ and $S$ exists.
Is this argument valid? If so, could it also be used to show no homomorphism exists between these two rings? If not, how would one prove the non-existence of a homomorphism?
This is false.
Define the map $\rho(a_0, a_1, \cdots) = (b_1, b_2, \cdots)$ such that $b_{2^i} = a_i$ and $b_j=0$ if $j$ is not a power of $2$. It can be shown that $\rho$ is an injective homomorphism from $R$ to $S$.
It was unclear what kind of sequences $A$ and $B$ you have in mind, but this doesn't matter. Say they are all from the same base ring $k$.
To shed some light, $R\simeq k[[x]]=\{\sum_{n=0}^\infty a_n x^n\}$, the formal power series ring which can be regarded as a completion of the semi-group algebra for the monoid $(\mathbb N, +)$, while $S\simeq \{\sum_{n=1}^\infty \frac{b_n}{n^s}\}$, the ring of formal Dirichlet series over $k$, or some completion of the semi-group algebra for the monoid $(\mathbb N^*, \cdot)$. Now it should be clear why there are homomorphisms from $R$ to $S$, because there are homomorphisms from $(\mathbb N, +)$ to $(\mathbb N^*, \cdot)$, such as $n\mapsto 2^n$. And it should also be very clear why $R$ and $S$ are both commutative, local, etc.
Similarly, we can have a nontrivial homomorphism from $S$ to $R$. List the prime numbers $p_1=2, p_2=3, p3, \cdots$, then we define the homomorphism $\rho(\prod_i p_i^{n_i})=\sum_in_i i$, which induces a homomorphism from $S$ to $R$. This essentially took advantage of the fact that $(\mathbb N^*, \cdot)$ is a free monoid with countably many generators (the prime numbers).