Printing all factorizations of a number
It is not immediately obvious that getPossibleProducts2() finds all
factorizations  and getPossibleProducts() removes the duplicates,
so the names could be chosen better. (And I would use "factorizations"
instead of "possibleProducts").

Your code is ineffective for two reasons:


The first method finds all factorizations with all possible permutations
(e.g. \$ 120 = 2 \cdot 60 = 60 \cdot 2 \$) so that you have to find the duplicates
in a second step.

As an example, for \$ n = 1024 \$, the first method computes
364176 factorizations, which are then reduced to 627 unique
factorizations in the second step.
You use strings to represents the factorizations (e.g. "2x60", "60x2") which then have to be split into integer arrays again.


To find duplicate factorizations, you split the string (e.g. "60x2")
to an integer array ["60", "2"], sort the factors (["2", "60"]) then concatenate the factors again ("260"). This is used as a hash key
to eliminate the duplicates. 
This is quite computing intensive. In particular the hash key

Arrays.stream(y).reduce("", (a, b) -&gt; a + b))


is compute twice, and StringBuilder might be a more effective way
to concatenate the array elements.

More important, it does not work: For \$ n = 130 \$,
the factorizations \$ 2 \cdot 65 \$ and  \$ 5 \cdot 26 \$ are both
mapped to the same hash key "265", so that the output of your
program is


[2, 65]
[13, 2, 5]
[10, 13]
[130]


without the factorization [5, 26].

Generally a better method would be to determine the factorizations
without duplicates in the first step. This can be done recursively
as follows:


Find the smallest factor f of the given number n.
Compute all factorizations of the n / f where the smallest
factor is at least f.


This automatically leads to factorizations with the factors in
increasing order and without duplicates.

As a side effect, the isPrime() function is not needed.

Here is a possible implementation. (Please take this as a demonstration of the algorithm 
only. Java is not my first language, so there may be more idiomatic
ways to do implement this.)

public static List&lt;int[]&gt; factorizations(int n) {
    // Start with a minimum factor of 2:
    return factorizations(n, 2, new ArrayList&lt;Integer&gt;());
}

public static List&lt;int[]&gt; factorizations(int n, int minFactor, ArrayList&lt;Integer&gt; previousFactors) {
    List&lt;int[]&gt; result = new ArrayList&lt;int[]&gt;();

    if (n == 1) {
        // This is where the recursion terminates. Convert the list of
        // all previously found factors to an int[] array
        // (found here: http://stackoverflow.com/a/23945015/1187415).
        result.add(previousFactors.stream().mapToInt(i-&gt;i).toArray());
    } else {
        for (int factor = minFactor; factor &lt;= n; factor++) {
            if (n % factor == 0) {
                // `factor` is a factor of `n`. Append it to (a clone of)
                // the previously found factors ...
                ArrayList&lt;Integer&gt; factors = new ArrayList&lt;Integer&gt;(previousFactors);
                factors.add(factor);

                // ... and (recursively) find all factorizations of `n / factor`
                // with factors greater or equal to `factor`:
                result.addAll( factorizations(n / factor, factor, factors) );
            }
        }
    }
    return result;
}


Example:

for (int[] f : factorizations(130)) {
    System.out.println(Arrays.toString(f));
}


produces the output


[2, 5, 13]
[2, 65]
[5, 26]
[10, 13]
[130]