Consider the following statement:
If $(x_i)_{i\in \Lambda}$ is a basis for the free $R$-module $M$, then $$ M=\bigoplus_{i \in \Lambda}Rx_i$$ My proof:
Clearly $\sum_{i \in \Lambda}Rx_i \subseteq M$. Now, by definition of basis, M is generated by $\{x_i|i \in \Lambda\}$, meaning that $M$ is the smallest $R$-submodule of itself containing each $x_i, i \in \Lambda$. However, if $1 \in R$, then we also have $x_i\in Rx_i$ for each $i \in \Lambda$, so that $\sum_{i \in \Lambda} Rx_i$ is an $R$-submodule of $M$ containing each $x_i, i \in \Lambda.$ Thus we also have $M \subseteq \sum_{i \in \Lambda}Rx_i$, so $M=\sum_{i \in \Lambda}Rx_i$. To show that this sum is direct, suppose $y \in \sum_{i \in \Lambda}Rx_i$ is expressible in two ways (as finite sums):
$$y=\sum_{i\in \Lambda}a_ix_i=\sum_{i\in \Lambda}b_ix_i$$
Then $\sum_{i \in \Lambda}(a_i-b_i)x_i=0_M$, and by the definition $(x_i)_{i \in \Lambda}$ is a basis, so $a_i=b_i$ for all $i$. Thus: The sum is direct. $\square$
This proof depends on the fact that $1 \in R$. The question is: If $R$ is a non-unital ring, is the statement still valid? Also, is the concept of basis (and linear independence of a subset) of a module over a ring still well defined / meaningful if $R$ does not necessarily have unity?
Given the definition you gave in the comments, the answer is no : take the nonunital ring $2\mathbb Z$ and the module $M=\mathbb Z$, with action indiced by the restriction to $2\mathbb Z$ of the usual one.
Then $(1)$ is a basis (it generates the module because any submodule containing it contains $1$ and $2k$ for any $k$, hence $2k+1$; and it is free for obvious reasons), but of course $\mathbb Z$ isn't equal to $2\mathbb Z \cdot 1$.
In this example however, there is an isomorphism between the two. An example where that doesn't happen is $\mathbb{2Z/4Z}$ acting on $\mathbb{Z/4Z}$. Then $1$ generates it for ibvious reasons, and it is free also for obvious reasons.
However, they don't have the same cardinality so they can't even be isomorphic.