A common "trick" for obtaining a closed form of a geometric series is to define $$ R := \sum_{k=0}^{\infty} r^k, $$ then manipulate the series as follows: \begin{align} R - rR &= \sum_{k=0}^{\infty} r^{k} + \sum_{k=0}^{\infty} r^{k+1} \\ &= (1 + r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\ &= 1 + (r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\ &= 1. \end{align} On the other hand, $R-rR = (1-r)R$. Hence $$ (1-r)R = 1 \implies R = \frac{1}{1-r}. $$ In this example, the formula is obtained by a sequence of relatively elementary algebraic manipulations.
By a similar kind of manipulation, suppose that $$ S := 1 + 1 + 1 + 1 + \dotsb = \sum_{k=0}^{\infty} 1. $$ $S$ is unaffected by addition of $1$, and so $S = 1+S$. Canceling $S$ from both sides gives $0 = 1$, which is clearly nonsense.
Question: What went wrong with the second computation? Why do these arguments work well for summing the geometric series, but not for the series of ones?
Definitions
The basic problem is that the object $S$ defined in the question is nonsense, at least within the scope of "normal" mathematical discourse. Thus the question really comes down to "Why is $R$ well-defined while $S$ is not?" The answer to this question comes down to definitions.
The Geometric Series
When working with a geometric series, we can obtain the result in the question directly from the definition. In that case, the partial sums are given by $$ R_n = \sum_{k=0}^{n} r^k. $$ The goal is to compute $\lim_{n\to\infty} R_n$, which can be done by first finding a useful closed form for each $R_n$. This can be done by paralleling the computations in the question, but in a way that can be justified rigorously. For each fixed $n$, we have \begin{align} (1-r)R_n &= R_n - rR_n \\ &= \sum_{k=0}^{n} r^k - \sum_{k=0}^{n} r^{k+1} \\ &= (1 + r + r^2 + \dotsb + r^{n-1} + r^n) - (r + r^2 + r^3 + \dotsb + r^n + r^{n+1}) \tag{1} \\ &= 1 + (r + r^2 + \dotsb + r^n) - (r + r^2 + \dotsb + r^n) - r^{n+1} \tag{2} \\ &= 1 + r^{n+1}. \end{align} At (1), we are just expanding out the notation. At (2), we are using the fact that addition is associative, and so we can move the parentheses around at will.[1] This computation then gives $$ (1-r)R_n = 1 - r^{n+1} \implies R_n = \frac{1-r^{n+1}}{1-r}. $$ As long as $r \ne 1$, this formula for the $n$-th partial sum is perfectly well-defined. If $r = -1$, then this expression oscillates between $\frac{1}{2}$ and $-\frac{1}{2}$, depending on the parity of $n$. Finally, if $|r| > 1$, then the magnitude of the numerator grows without bound, and the sequence of partial sums fails to converge. Otherwise, i.e. if $|r| < 1$, we can take a limit to get $$ \lim_{n\to\infty} R_n = \lim_{n\to\infty} \frac{1-r^{n+1}}{1-r} = \frac{1}{1-r}. $$ Therefore, from the definition of a convergent series, we are justified in writing $$ \sum_{k=0}^{\infty} r^k = \frac{1}{1-r}, $$ assuming that $|r| < 1$. The "algebraic manipulations" in the question are, in a sense, a shortcut through this more formal computation.
The Series of Ones
In the case of the series of ones, things go wrong. The $n$-th partial sum is given by $$ S_n = \sum_{k=0}^{n} 1 = n+1. $$ But then $$ \lim_{n\to\infty} S_n = \lim_{n\to\infty} (n+1) = \infty. $$ The sequence of partial sums is unbounded, and therefore does not converge to a real number. In other words, the series $$ \sum_{k=0}^{\infty} 1 = 1+1+1+1+\dotsb $$ cannot reasonably be assigned a real value. Since it cannot be assigned a real value, further algebraic manipulation is meaningless. Further discussion of this problem can be found in the answers to this question about arithmetic with infinite quantities.
[1] Since we are working with infinite series here, it is important to note that finite addition is associative. That is, if we have a finite number of terms which we want to add together, we can rearrange the parentheses however we like. This does not work with an infinite number of terms. See, for example, Grandi's series.