For any $n\times n$ matrices $X_1$ and $X_2$, \begin{equation} \begin{aligned} e^{X_1+X_2}-e^{X_1}&=\sum_{k=0}^\infty\frac1{k!}\,[(X_1+X_2)^k-X_1^k] \\ &=\sum_{k=0}^\infty\frac1{k!}\; \sum_{(j_1,\dots,j_k)\in\{1,2\}^k\setminus\{(1,\dots,1)\}}\; \prod_{i=1}^k X_{j_i}, \end{aligned} \end{equation} so that \begin{equation} \begin{aligned} \|e^{X_1+X_2}-e^{X_1}\| &\le\sum_{k=0}^\infty\frac1{k!}\; \sum_{(j_1,\dots,j_k)\in\{1,2\}^k\setminus\{(1,\dots,1)\}}\; \prod_{i=1}^k \|X_{j_i}\| \\ &=\sum_{k=0}^\infty\frac1{k!}\; \sum_{m=1}^k \binom km \|X_1\|^m \|X_2\|^{k-m} \\ &=\sum_{k=0}^\infty\frac1{k!}\;[(\|X_1\|+\|X_2\|)^k-\|X_1\|^k] \\ &=e^{\|X_1\|+\|X_2\|}-e^{\|X_1\|} \\ &\le \|X_2\| e^{\|X_1\|+\|X_2\|}.\quad\Box \end{aligned} \end{equation}

For any $n\times n$ matrices $X_1$ and $X_2$, \begin{equation} \begin{aligned} e^{X_1+X_2}-e^{X_1}&=\sum_{k=0}^\infty\frac1{k!}\,[(X_1+X_2)^k-X_1^k] \\ &=\sum_{k=0}^\infty\frac1{k!}\; \sum_{(j_1,\dots,j_k)\in\{1,2\}^k\setminus\{(1,\dots,1)\}}\; \prod_{i=1}^k X_{j_i}, \end{aligned} \end{equation} so that (assuming that the matrix norm is an operator norm) \begin{equation} \begin{aligned} \|e^{X_1+X_2}-e^{X_1}\| &\le\sum_{k=0}^\infty\frac1{k!}\; \sum_{(j_1,\dots,j_k)\in\{1,2\}^k\setminus\{(1,\dots,1)\}}\; \prod_{i=1}^k \|X_{j_i}\| \\ &=\sum_{k=0}^\infty\frac1{k!}\; \sum_{m=1}^k \binom km \|X_1\|^m \|X_2\|^{k-m} \\ &=\sum_{k=0}^\infty\frac1{k!}\;[(\|X_1\|+\|X_2\|)^k-\|X_1\|^k] \\ &=e^{\|X_1\|+\|X_2\|}-e^{\|X_1\|} \\ &\le \|X_2\| e^{\|X_1\|+\|X_2\|}.\quad\Box \end{aligned} \end{equation}

As shown in Christian Remling's answer, the condition that the matrix norm is an operator norm cannot be dropped.