This is not really an answer to this question as posted (I do not discuss the exponential function), but it points out that a certain well-known method might be interpreted as induction over the reals. The idea is that to prove something is true, for say all positive $x$, one could prove it for $x=0$ and then show that the set $T\subseteq\mathbb R$ for which it is true is both open and closed.

Compared to usual induction, when we would like to prove a statement or property $P(n)$, we prove $P(0)$ and then (induction step), show that there is no largest integer $n$ for which the statement is true. Indeed, if $n$ were largest, then $P(n)$ is true, but then $P(n+1)$ must also be true, contradicting that $n$ was largest.

Back to the case of the reals, if $T$ had a largest element $t$, then since $T$ is open we have $t+\varepsilon\in T$ for all small enough $\varepsilon>0$, contradicting that $t$ were largest.
Note that $T$ must also be unbounded, since if $\sup T<\infty$ then using that $T$ is closed we would have $\sup T\in T$ so $\sup T$ would be the largest element of $T$, but we already saw that $T$ cannot have a largest element.

The above explains the idea but does not yet constitute a proof that $[0,\infty)\subseteq T$. To make it precise recall that a version of induction for the integers is the so-called complete induction, where (after starting with $P(0)$) one proves that $P(m+1)$ holds whenever $P(n)$ holds for all $n$ in the interval of integers $[0,m]$ (that is, whenever $P(n)$ holds for all $n\in\{0,1,\dots,m\}$). (Restated in a different way: One proves that $P(n+1)$ holds whenever each of $P(0),P(1),\dots,P(n)$ holds.)

For the reals, we could define $s=\sup\{x\ge0:[0,x]\subseteq T\}$, and then we would like to show that $s=\infty$.
But if $s<\infty$ then, since $T$ is closed we have $s\in T$, and since $T$ is open we have $s+\varepsilon\in T$ for all small enough $\varepsilon>0$, contradicting the definition of $s$.

To illustrate the above with a real-world-example, my answer was inspired by the following MSE question by Saun Dev and more specifically by this answer by Michael Burr to (for which the OP requested an elaboration ... my answer here may provide this).

So the problem there was to prove that if
$f:\mathbb R \to \mathbb R$ is a differentiable function such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ then $f(x)>0,\forall x>0$. The answer I referred to provided the following hint: Consider $x=0$. Since $f'(0)>f(0)=0$, we know
$$
\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0}\frac{f(h)}{h}>0.
$$
Therefore, for $h$ positive and sufficiently close to $0$, the numerator is positive. Continue with this idea.

So to elaborate, let $T(x)$ be the statement that either $x=0$ or $x>0$ and $f(x)>0$. (Formally $T(x)$ is $(x=0)\lor((x>0)\land(f(x)>0))$.)
Let $T=\{x:T(x)\}$. What the hint shows is that $T$ is (relatively) open at $0$ in the interval $[0,\infty)$, that is there is $\varepsilon>0$ such that $T(h)$ holds, whenever $h\in[0,\varepsilon)$. Namely $T(h)$ holds for $h\in(0,\varepsilon)$ since $f(h)>0$ (by the hint), and $T(0)$ holds (trivially).

The *induction step* here could be split into two parts. First, $T$ is open, if $x\in T$ then *we could push it further to the right*, $T$ is open, there is $\varepsilon>0$ such that $T(h)$ holds, whenever $h\in[x,x+\varepsilon)$. This is essentially what the hint shows. Using that $f'(x)>f(x)\ge0$ (for $x\ge0$) and
$$
\lim_{h\rightarrow x}\frac{f(x+h)-f(x)}{h}=f'(x)>0.
$$
we conclude that $f(x+h)>f(x)\ge0$ for $h\in[x,x+\varepsilon)$
(or even for $h\in(x-\varepsilon,x+\varepsilon)$, if $x>0$) for all sufficiently small $\varepsilon>0$.

Second part of the *induction step* (which resembles complete induction for the integers), if $x>0$ and $T(y)$ holds for all $y\in[0,x)$ then $T(x)$ holds too. Indeed in the example here, since $f'(y)>f(y)>0$ for all $y\in(0,x)$, then $f$ must be strictly increasing on $(0,x)$ and by continuity $f(x)>0$ (and, with more detail, $f(x)>f(y)>0$ for each $y\in(0,x)$).