Lets say that a commutative semiring $S$ has the unique basis property iff each $S$-module has at most one basis. For example, both $\mathbb{N}$ and $\mathbb{B}$ have the unique basis property, where $\mathbb{B} = \{0,1\}$ is the boolean domain, i.e. the quotient of $\mathbb{N}$ by the relation $1+1=1$.
Note that the trivial ring $T = \{0\}$ does not have the unique basis property, because both $\emptyset$ and $\{0\}$ are bases for $T$. More generally, no ring has this property. For suppose $R$ is an arbitrary commutative ring. Then both $\{(1,0),(0,1)\}$ and $\{(1,1),(0,1)\}$ are bases of $R^2$. To see that the latter is indeed a basis, firstly note that since $(a,b) = a(1,1)+(b-a)(0,1),$ hence this is indeed a generating set. Secondly, assume that $$A(1,1)+B(0,1)=0.$$ Then $A=0$, so we get $B(0,1)=0$, which implies that $B$ equals $0$. So $\{(1,1),(0,1)\}$ is always a basis of $R^2$, for any ring $R$.
Following this line of thought, we might ask the following:
Preliminary Question. Suppose $S$ is a non-trivial semiring in which no element but $0$ has an additive inverse. i.e., we're assuming $$a+b=0 \rightarrow a = 0 \wedge b = 0.$$ Does this imply that $S$ has the unique basis property?
But the answer is an obvious no: take $\mathbb{R}_{\geq 0}$, for example. The problem is basically that for each unit $u \in S$, the set $\{u\}$ is a basis of $S$. Okay, so lets do the obvious thing by omitting all such examples:
Question. Suppose $S$ is a non-trivial semiring in which no element but $0$ has an additive inverse, and the only unit of $S$ is $1$. Does $S$ necessarily have the unique basis property?