How to conclude that for all $f,g \in S_I$, we have that if $f(x)=g(x)$ for some $x \in I,$ then $f=g$?

How to conclude that for all $f,g \in S_I$, we have that if $f(x)=g(x)$
for some $x \in I,$ then $f=g$?

Motivation. If we're trying to solve the differential equation $y'=y$ on
the interval $I$, we would probably like to assume that for all $x \in I$
we have that $y(x) \neq 0.$ But, we need to justify this assumption by
showing that, if we assume to the contrary that there exists $x \in I$
such that $y(x) = 0,$ then the only possible solution is $y=0_I$, where
$0_I$ is the function with domain $I$ that is everywhere $0$. This
motivates the following question.
Question. Suppose that for all sets $X \subseteq \mathbb{R},$ we define
that $$S_X = \{y : X \rightarrow \mathbb{R} \mid y'=y;\; y \mbox{
diff}\}.$$
Under this definition, is there some theorem that allows us to conclude
that for all non-trivial* intervals $I$ and for any two $f,g \in S_I$, we
have that if $f(x)=g(x)$ for some $x \in I,$ then $f=g$?
*By a non-trivial interval, I mean an interval incorporating two or more
elements (which implies that the interval has infinitely many elements).
Actually, I'm not certain whether this assumption is strictly necessary,
but it seems harmless.