No, that is not possible.
Assume that player $2$ has a strategy $s_{2j'}$ that is strictly dominated by $s_{2j}$. This means that for all pure strategies $s_1$ of player $1$, we have
$$
f_2(s_1,s_{2j}) > f_2(s_1,s_{2j'}).
$$
Lemma 1.
This above inequality is also true for any mixed strategies $s_1$ of player $1$.
Proof. Assume player $1$ mixes with a probability vector $p$. Thus the expected payoffs of player $2$ when playing strategies $s_{2j}$ and $s_{2j'}$ are respectively
$$
\sum_i p_i \cdot f_2(s_{1i},s_{2j}), \hskip 10pt \text{and} \hskip 10pt \sum_i p_i \cdot f_2(s_{1i},s_{2j'}).
$$
For positive probabilities $p_i$ any member $p_i \cdot f_2(s_{1i},s_{2j})$ of the left hand sum will be larger than the corresponding $p_i \cdot f_2(s_{1i},s_{2j'})$ member of the right hand sum. When $p_i = 0$ the members too will be equal. $\sum_i p_i = 1$, so at least one $p_i$ is positive, thus the left hand sum is itself larger than the right hand sum; hence $s_{2j}$ dominated $s_{2j'}$ even against mixed strategies. $QED$
Lemma 2.
If a mixed strategy $s_2'$ of player $2$ places positive weight $q_{j'}$ on her pure strategy $s_{2j'}$, then player $2$ has a mixed strategy $s_2$ that dominates $s_2'$.
Proof. Player $2$ can increase her payoff by removing the probability from $s_{2j'}$, because regardless of the strategy $s_1$ player $1$ plays we have
$$
\begin{align*}
f_2(s_{1},s_{2j'}) & < f_2(s_{1},s_{2j}) \\
q_{j'} \cdot f_2(s_{1},s_{2j'}) & < q_{j'} \cdot f_2(s_{1},s_{2j}) \\
q_{j'} \cdot f_2(s_{1},s_{2j'}) + \sum_{i\neq j'} q_i \cdot f_2(s_{1},s_{2i}) & < q_{j'} \cdot f_2(s_{1},s_{2j}) + \sum_{i\neq j'} q_i \cdot f_2(s_{1},s_{2i}).
\end{align*}
$$
The left hand side of the final inequality is the expected playoff of player $2$ while playing $s_2'$, while the larger right hand side is her expected payoff while playing another strategy. This other strategy - which we will denote by $s_2$ - yields a larger payoff than $s_2'$ irrespective of $s_1$, thus $s_2$ strictly dominated $s_2'$. $QED$
One can now perform the iteration of strictly dominated pure strategies again. In the end a player should never put positive weight on a strictly dominated pure strategy, even when facing mixed strategies. In equilibrium players act optimally, they will place 0 positive weight on all eliminated strategies. Thus in a dominance solvable game in equilibrium players will play pure strategies.
