In Lee's Introduction to Smooth Manifolds he states:
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, we define a map $\exp: \mathfrak{g} \rightarrow G$, called the exponential map of $G$, as follows: for any $X \in \mathfrak{g}$, we set $$\exp X = \gamma(1),$$ where $\gamma$ is the one-parameter subgroup generated by $X$, or equivalently the integral curve of $X$ starting at the identity.
Why do we define $\exp X$ as $\gamma(1)$? What is special about the value of $\gamma$ at $1$?
The basic idea is that we want to travel in the direction of $x$ for a single unit of time to see where it takes us.
Think about the simplest example where $G$ is the multiplicative group of positive reals and $\mathfrak g$ is the additive reals.
Like most things in mathematics, the definition is usually motivated by its properties. In particular here, the $1$ is nice since it effectively causes the differential of this map to just be the identity. In other words, if we think as Lie algebras as the behavior of the group near the identity, then the 1 cancels out so the exponential map acts as the identity near 1.
Then if you choose some positive real $x$, this defines some curve (i.e. one parameter subgroup of G) with initial velocity $x$ as $\gamma(t)=e^{tx}$. This gives that $\exp x= \gamma(1)=e^x$. In other words, the exponential map of the the Lie group of the multiplicative reals matches our standard exponential function.
One possible objection is that if you’re familiar with physics dimensional analysis, then it feels a little weird to have something happen at 1 second because we might change our unit of time which could change the definition. However, we’ve defined the velocity of this path as $x$ distance per 1 “second”, so the velocity and the time cancel out when computing your location after that unit of time.