A function $f(x)$ analytic at $x=0$ can be represented as power series within an open disc with radius of convergence $R$.
\begin{align*}
f(x)=\sum_{n=0}^\infty a_nx^n\qquad\qquad \qquad |x|<R
\end{align*}
Any substitution $x=g(u)$ is admissible as long as we respect the radius of convergence.
\begin{align*}
f(g(u))=\sum_{n=0}^\infty a_n \left(g(u)\right)^n\qquad\qquad\quad |g(u)|<R
\end{align*}
We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$.
\begin{align*}
f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R}
\end{align*}
Substitution $u=2x$
We consider
\begin{align*}
f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad 2x\in \mathbb{R}
\end{align*}
This substitution is admissible for all $x \in \mathbb{R}$ since $$2x\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(2x)=e^{2x}$ is $\infty$.
We obtain
\begin{align*}
f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R}
\end{align*}
Substitution $u=x+1$
We consider
\begin{align*}
f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x+1\in \mathbb{R}
\end{align*}
This substitution is admissible for all $x \in \mathbb{R}$ since $$x+1\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(x+1)=e^{x+1}$ is $\infty$.
We obtain
\begin{align*}
f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R}
\end{align*}
We also obtain
\begin{align*}
e\cdot e^x&=\left(\sum_{k=0}^\infty \frac{1}{k!}\right)\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{k!}\cdot\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^n\frac{1}{(n-l)!}\cdot\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{l=0}^n\binom{n}{l}x^l\right)\frac{1}{n!}\\
&=\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\\
&=e^{x+1}
\end{align*}
Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.
