When adding summations, for example, $\sum_{n = 1}^{\infty} 3x^{n - 1} + \sum_{n = 1}^{\infty} nx^n$, a variable change such as $m = n - 1$ is often made, in this case yielding $\sum_{m = 0}^{\infty} 3x^m + \sum_{n = 1}^{\infty} nx^n$. Although the motivation to do so depends on this question having been answered, I'll break a term out of the first summation to match the indices, yielding $3 + \sum_{m = 1}^{\infty} 3x^m + \sum_{n = 1}^{\infty} nx^n$.
The next step is to rename $m$ as $n$, and proceed to combine and factor the expression into $3 + \sum_{n = 1}^{\infty} (3 + n)x^n$. This is the part that seems very unnatural to me. The explanations I've heard to justify renaming $m$ as $n$ are "What does the name of a variable really matter? It's just a name!" and "$m$ and $n$ are just iteration variables that won't be present in the final expression." Neither of these is very satisfying. I get that it doesn't matter what name a variable has. There is a very natural and obvious bijection between $r + 2 = 6$ and $y + 2 = 6$. I even get that a variable name can be used multiple times in a problem. You can use $A$ in an ODE system to represent both the coefficient matrix and an arbitrary constant, provided you keep the uses straight in your mind. The problem here is that factoring step doesn't seem to keep the uses straight. It treats old $n$ and new $n$ as the same variable. The fact that $m$ and $n$ are iteration variables also doesn't seem to imply that they're the same variable; if values of $n$ represent days in a week ($n = 1$ is Sunday, $n = 2$ is Monday, etc.), then equivalent values of $m$ (or new $n$) represent different days (shifted by $1$).
I am not looking for an answer that expands a summation and its shifted version to prove that they match. Also, I can do the algebra for a summation shift quickly with no problem, and it makes perfect sense at that level of abstraction. It's only when I slow down and actually make the variable changes that things start to feel wrong.
This is really just explaining Greg Martin's comment:
Let $f$ and $g$ be functions, and for each $n$, let $A_n={\displaystyle \sum_{a=1}^n} f(a)$, $B_n={\displaystyle \sum_{b=1}^n} g(b)$, and $C_n={\displaystyle \sum_{c=1}^n} \left(f(c)+g(c)\right)$.
Note that $C_n=(f(1)+g(1))+(f(2)+g(2))+\cdots+(f(n)+g(n))$, so that by associativity and commutativity of (finite) addition, $C_n=A_n+B_n$. Then by limit laws, ${\displaystyle \lim_{n\to\infty}}A_n+{\displaystyle \lim_{n\to\infty}}B_n={\displaystyle \lim_{n\to\infty}}C_n$ whenever you know at least two of the limits exist (it's a sum law if you know the $A$ and $B$ limits exist, and it's a difference law otherwise).
However, by the definition of the sum of a series, this means that we have
$$\boxed{{\displaystyle \sum_{a=1}^\infty} f(a)+{\displaystyle \sum_{b=1}^\infty} g(b)={\displaystyle \sum_{c=1}^\infty} \left(f(c)+g(c)\right)\text{if two series converge}}\tag{$1$}$$
Writing $a$ as $m$ and $b$ as $n$ and $c$ as $n$, but keeping the uses straight (each $n$ appears in its own sum on its own side of the equation), we obtain
$$\boxed{{\displaystyle \sum_{m=1}^\infty} f(m)+{\displaystyle \sum_{n=1}^\infty} g(n)={\displaystyle \sum_{n=1}^\infty} \left(f(n)+g(n)\right)\text{if two series converge}}\tag{$2$}$$
But this says exactly that the step of "rename $m$ to $n$ and then combine the two series, treating the two $n$s the same" is mathematically valid*.
*But this is only guaranteed to work under the assumption that either: 1. both of the original $m$ and $n$ series converge. or 2. one of the $m$ and $n$ series converges, and the end result converges too.
For example, if we choose $f(m)=\dfrac{1}{2m-1}$ and $g(n)=-\dfrac{1}{2n}$, then the left side is undefined (it looks like the indeterminate form $\infty-\infty$) but the right side is $\ln2$.
For this sort of reason, you were right to be cautious. But, in practice, this issue doesn't come up often in calculations.