I encourage you to work through the details of your own approach. Here is my alternative.
After the first drawing (a pair of balls), we have exactly two balls (out of five possible) that have been drawn (and we can never get fewer than this in the end).
The second drawing can affect which balls have been seen. There are three possible outcomes as far as the number of different balls drawn in either the first or second drawing.
0. We might get the same two balls as we did the first time.
1. We might get one of the original two balls and a third ball not seen in the first drawing.
2. We might get neither of the original two balls, so two new balls from this second drawing (for a total of four seen after both drawings).
Given that there are $\binom{5}{2}=10$ ways the second drawing can occur, counting the number of $2$-subsets of the five (distinct) balls, it should be pretty easy to work out the probability for each of these three outcomes. Only one of the ten ways corresponds to case 0. Case 2. corresponds to $\binom{3}{2}=3$ of them. Thus case 1. has probability $6$ out of ten, or $0.6$.
If case 0. occurred in the second drawing, nothing that happens in the third and final drawing will result in getting all five balls. We will fail to get all five balls at least once.
If case 1. occurred in the second drawing, then we have a "fighting chance". We have seen three of the balls, and what is needed is to draw exactly the two balls in the third drawing that had not been seen before. As just pointed out, the chance of doing that in the third drawing is the same as case 0. in the second drawing: one chance in ten.
Finally if case 2. occurred in the third drawing, it will be a lot more likely to succeed in getting the one missing ball (we have two "chances" to get it). A little calculation shows that the probability of going from case 2. in the second drawing to getting all five balls in the end is: four chances in ten. (It might be easier to count how many ways there are to fail to get the last ball, namely six ways out of ten.)
Now put the possible paths to success together in a non-overlapping way. We could get all five balls in three drawings either by:
A. Getting case 1. in the second drawing and filling out the sample in the third drawing. Probability is $0.6 \times 0.1 = 0.06$.
B. Getting case 2. in the second drawing and sampling the missing fifth ball in the third drawing. Probability is $0.3 \times 0.4 = 0.12$.
Add the probabilities of these disjoint outcomes and you have the combined probability that all five balls will appear at least once in the three samples.
Let me outline how we can make this computation scale up to similar but more complicated situations. In the Markov chain approach, which we've sketched in words above, one identifies "states" that occur and the probabilities (when a drawing is made) of transitioning from one state to another. We then format all these probabilities into a state-transition matrix.
Here we have considered states based on how many balls have been seen. Originally no balls were yet seen, and it is possible that up to five balls will be seen (after three draws). So we'll label the six states $B_0,B_1,\ldots,B_5$ according to how many different balls have been seen.
The Reader will note that when a draw begins in state $B_0$, there is a probability of $1$ that a transition to state $B_2$ will occur. Similarly the transition probabilities from state $B_2$ are calculated above as follows:
$$ \begin{align*}
B_2 &\to B_0 &: \; 0.0 \\
&\to B_1 &: \; 0.0 \\
&\to B_2 &: \; 0.1 \\
&\to B_3 &: \; 0.6 \\
&\to B_4 &: \; 0.3 \\
&\to B_5 &: \; 0.0
\end{align*} $$
If we compile all the transition probabilities into a matrix, where $M_{ij}$ is the chance of going from $B_i$ to $B_j$, then:
$$ M = \begin{bmatrix}
0 & 0 & 1.0 & 0 & 0 & 0 \\
0 & 0 & 0.4 & 0.6 & 0 & 0 \\
0 & 0 & 0.1 & 0.6 & 0.3 & 0 \\
0 & 0 & 0 & 0.3 & 0.6 & 0.1 \\
0 & 0 & 0 & 0 & 0.6 & 0.4 \\
0 & 0 & 0 & 0 & 0 & 1.0 \end{bmatrix} $$
Letting $p_0,p_1,\ldots,p_5$ be the probabilities of states before a draw and $q_0,q_1,\ldots,q_5$ the probabilities after a draw. Then:
$$ \begin{bmatrix} p_0 & p_1 & p_2 & p_3 & p_4 & p_5 \end{bmatrix}
M = \begin{bmatrix} q_0 & q_1 & q_2 & q_3 & q_4 & q_5 \end{bmatrix} $$
Given that the initial "probability" distribution of states (before any drawing) is $p_0 = 1$ and the rest zeros, it is not terribly hard to show that the distribution after three drawings is:
$$ \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}
M^3 = \begin{bmatrix} 0.00 & 0.00 & 0.01 & 0.24 & 0.57 & 0.18 \end{bmatrix} $$
The final entry of this last probability distribution of states is our answer.
