This follows from a result of Klyachko. Klyachko proved:

Let $\alpha$, $\beta$ and $\gamma$ be three vectors in $\mathbb{R}^n$. Then the following are equivalent:

(1) There exist Hermitian matrices $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with $\mathfrak{a}+\mathfrak{b}= \mathfrak{c}$, and with eigenvalues $\alpha$, $\beta$ and $\gamma$ respectively.

(2) There exist invertible matrices $A$, $B$ and $C$ with $AB=C$ and singular values $e^{\alpha}$, $e^{\beta}$ and $e^{\gamma}$ respectively. Here $e^{\alpha}$ etcetera mean termwise exponentiation.

I will show that $(2) \implies (1)$ implies your statement (and is basically equivalent to it). I'll call your Hermitian matrices $X$ and $Y$, to leave the letters $(A,B,C)$ clear.

Let $A=e^{X/2}$, let $B=e^{Y/2}$ and let $AB=C$. Let $e^{\alpha}$, $e^{\beta}$ and $e^{\gamma}$ be the singular values of $A$, $B$ and $C$. So the eigenvalues of $C C^{\ast}$ are $e^{2 \gamma}$, and we note that $C C^{\ast} = e^{X/2} e^Y e^{X/2}$.

Using $(2) \implies (1)$, let's find Hermitian $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with eigenvalues $2 \alpha$, $2 \beta$ and $2 \gamma$ and $\mathfrak{a}+\mathfrak{b} = \mathfrak{c}$. Then $C C^{\ast}$ and $e^{\mathfrak{c}}$ are Hermitian with the same eigenvalues and, conjugating by a unitary matrix, we can arrange that $C C^{\ast} = e^{\mathfrak{c}}$.

Now, $X$ and $\mathfrak{a}$ are both Hermitian with eigenvalues $2 \gamma$, so we can find unitary $U$ with $\mathfrak{a} = U X U^{\ast}$. Similarly, we can find unitary $V$ with $\mathfrak{b} = V Y V^{\ast}$. So $\mathfrak{c} = U X U^{\ast} + V Y V^{\ast}$ and $e^{X/2} e^Y e^{X/2} = e^{U X U^{\ast} + V Y V^{\ast}}$ as desired.