The answer seems to be no. Moreover: Suppose that every set of reals has the property of Baire. Let $\mathbb{C}$ be Cohen forcing and let $P$ be any wellorderable partial order. If $(c,d)$ is generic for $\mathbb{C} \times P$, then there is no injection in $V[c,d]$ from the Cantor space of $V[c]$ to any set in $V[d]$.
Here's a proof, which is a variation on the standard argument showing that if all sets of reals have the property of Baire, then there is no function choosing between complementary $\mathbb{E}_{0}$ degrees. We will call members of the Cantor space reals. Suppose that $\tau$ is a $\mathbb{C} \times P$ name for a function mapping the reals of $V[g_{left}]$ into $V[g_{right}]$, where $g_{left}$ and $g_{right}$ are the $\mathbb{C} \times P$-names for the left and right coordinates of the generic. We will find a condition forcing the realization of $\tau$ not to be injective.
Let $N$ be the set of nice Cohen names $\sigma$ for elements of the Cantor space which are also functions with domain $\mathbb{C}$, where each value $\sigma(p)$ has the form $\check{t}$, for $t$ an element of $\mathbb{C}$ of the same length as $p$. That is, $N$ is the set of nice $\mathbb{C}$-names for reals such that each condition of $\mathbb{C}$ decides the initial segment of the realization of the name up to the length of the condition (and maybe more).
Let $\langle q_n : n \in \omega \rangle$ be the natural enumeration of $\mathbb{C}$, where shorter sequences are listed before longer ones, and sequences of the same length are listed in lexicographic order (any enumeration in ordertype $\omega$ would work). Given $p \in \mathbb{C}$, let $n_p$ be such that $p = q_{n_{p}}$. There is a natural bijection $b$ between the Cantor space and $N$, where we let $x$ in the Cantor space correspond to the set of pairs $(p, \check{t})$, where $t = \langle x(n_{p \upharpoonright i}) : i \leq |p| \rangle$.
Applying the assumption that every subset of the Cantor space has the property of Baire, and the Kuratowski-Ulam theorem to deal with the wellorderable poset $P$, we get $s$ in $\mathbb{C}$ and $(p_0, p_1) \in \mathbb{C} \times P$ such that, for comeagerly many $x$ extending $s$, there is some $\mathbb{C}$-name $\rho$ with $(p_0,p_1)$ forcing $(b(x)_{g_{left}}, \rho_{g_{right}}) \in \tau$. That is, $(p_0,p_1)$ forces some $P$-name $\rho$ to represent via $g_{right}$ the $\tau$-value for the $g_{left}$-realization of $b(x)$. However, we can find $x$ and $x'$ extending s and in this comeager set such that $x$ and $x'$ disagree at exactly one value $n^*$ with $q_{n^*}$ compatible with $p_0$, and with this $q_{n^*}$ a proper extension of $p_0$. So $p_0$ doesn't decide whether or not the realizations of $b(x)$ and $b(x')$ will be the same.
Moreover, there exist $P$-names $\rho$ and $\rho'$ such that $(p_0,p_1)$ forces both
$(b(x)_{g_{left}}, \rho_{g_{right}}) \in \tau$
and
$(b(x')_{g_{left}}, \rho'_{g_{right}}) \in \tau$.
Let $p'_1$ be a strengthening of $p_1$ deciding whether or not $\rho$ and $\rho'$ have the same realization.
Since there are extensions of $p_0$ forcing that the realizations of $b(x)$ and $b(x')$ will be the same, and $\tau$ is a name for a function, it must be that $p'_1$ forces the realizations of $\rho$ and $\rho'$ to be the same. The condition $q_{n^*}$ however extends $p_0$ and forces that the realizations of $b(x)$ and $b(x')$ will be different, and therefore $(q_{n^{*}}, p'_1)$ forces that the realization of $\tau$ will not be an injection.
