For $0\leq x\in\mathbb{R}$, let q=$\lfloor{x}\rfloor\in\mathbb{N}$ be the greatest integer $\leq x$ and let $r=x-q\in[0,1)$ be the (nonnegative) fractional part of $x$, so that $x=q+r$. Then for $n\in\mathbb{N}$, $ |x-n|= \left\{ \begin{matrix} (q-n)+r &\quad n\leq q\leq x \\ (n-q)-r &\quad n>x>q \end{matrix} \right. $ so that $ \begin{array}{} \sum_{n=0}^{\infty}e^{-|x-n|} &=& \sum_{n=0}^{q}e^{-r-(q-n)} &+& \sum_{n=q+1}^{\infty}e^{+r-(n-q)} \\&=& e^{-r}\sum_{n=0}^{q}e^{-(q-n)} &+& e^r\sum_{n=q+1}^{\infty}e^{-(n-q)} \\&=& e^{-r}\sum_{n=0}^{q}e^{-n} &+& e^{r-1}\sum_{n=0}^{\infty}e^{-n} \\&=& e^{-r}\frac{1-e^{-(q+1)}}{1-e^{-1}} &+& e^{r-1}\frac{1}{1-e^{-1}} \\&=& \frac{e^{r}+e^{1-r}+e^{-x}}{e-1} && \end{array} $ converges absolutely (since each term is positive) -- but not uniformly! -- for all $x$.
(Aside:) If we extend the finite sum above (over $0\leq n\leq q$) to an infinite sum (over all $n\geq0$), we can derive a simple upper bound on the series of $\frac{2}{1-e^{-1}}$; in fact, the series sum always lies in $\left(\frac{2\sqrt e}{e-1},\frac{e+2}{e-1}\right]$.
Note, however, that in both cases, we have split the infinite sum on the left into two sums: a finite sum of increasing terms, stopping at $q$, the greatest integer less than or equal to $x$, and an infinite sum of decreasing terms. Both sums are geometric series and converge. The "cut" $q$, however, depends on $x$, and the largest term is either $e^{-(x-q)}=e^{-r}$ (when $r\leq\frac12$) or $e^{x-(q+1)}=e^{r-1}$ (when $r\geq\frac12$), which occurs twice in case $r=\frac12$. This largest term, then, depends on the distance $d(x)=\min(r,1-r)$ of $x$ from $\mathbb{Z}$ in $\mathbb{R}$, which always lies in $[0,\frac12)$, and so has value $e^{-d(x)}$ which, in particular, is always $\geq e^{-\frac12}$.
Consequently, the series is not uniformly convergent. According to the definition of uniform convergence (of a sequence of functions $f_n$ from some set to $\mathbb{R}$ or $\mathbb{C}$ or a metric space):
A sequence of functions $f_n:X\rightarrow\mathbb{R}$ is uniformly convergent if, firstly:
for each $x\in X$, $f_n(x)$ converges pointwise to a finite value $f(x)$, so that there exists a well-defined limit function $f:X\rightarrow\mathbb{R}$, in which case we say that $(f_n)$ converges and write $f_n\rightarrow f$,
and secondly, if:
for each $\epsilon>0$, there exists an $N\in\mathbb{N}$ so that for all $n \geq N$ (and all $x\in X$), we have $|f_n(x)-|f(x)|<\epsilon$.
In particular, $N=N(\epsilon)$ can depend on $\epsilon$ but not on $x\in X$. This is the sense of the word uniform.
To apply this definition to our series, we need to find the sequence of functions to which the definition applies. This will be the sequence $f_n$ of partial sums $ f_n(x)=\sum_{k=0}^{N}e^{-|x-k|} $ and to prove its uniform convergence, we would want to show that the differences $ |f_n(x)-f(x)|=\sum_{k=N+1}^{\infty}e^{-|x-k|} $ could be made smaller than any desired $\epsilon>0$, for all $x$, given a suitable choice of (just one value of) $N$. In our case, we can't provide an $N$ ahead of time, without foreknowledge of $x$, since the terms peak near $x$, as discussed above.
An equivalent formulation for uniform convergence is that the sequence $\alpha_n=\sup_{x\in X}|f_n(x)-f(x)|$ tends toward zero as $n\rightarrow\infty$.
However in our case, using $x=n+1$ to bound $\alpha_n$ below, $ \alpha_n =\sup_{x\geq0}\sum_{k=n+1}^{\infty}e^{-|x-k|} \geq\frac1{1-e^{-1}}>0 \quad\implies\quad \alpha_n\nrightarrow 0 \quad\text{as}\quad n\rightarrow\infty $ showing that the convergence is not uniform. (Aside: In fact, $\alpha_n=\frac{1+e^{-1}}{1-e^{-1}}=\frac{e+1}{e-1}$.)