If $‎\lim\limits_{x\to\infty}‎\frac{f(x)}{g(x)} = 1‎$,‎ then $\lim\limits_{x\to\infty}(f(x) - g(x)) = 0‎$.

You need $f(x) = g(x) + h(x)$ where both $h(x) = o(1)$ and $h(x) = o[g(x)]$.

... or stated more precisely by Mark Viola, these are necessary and sufficient conditions.

For example, if $f(x) = e^{-x}+ 1/x$ and $g(x)= e^{-x}$, then the second condition is violated. In your example, the first condition is violated.

The conditions (A) $\lim_{x\to\infty}f(x)/g(x)=1$ and (B) $\lim_{x\to\infty}[f(x)-g(x)]=0$ are pretty much unrelated. For example, $f(x):=2/x$ and $g(x):=1/x$ satisfy B but not A, while your example shows independence in the other direction. Any "additional" condition that implies B will need to be, essentially, a statement of B itself (or something stronger). Condition A won't be making a useful contribution.

Notice that from $\lim\limits_{x\to\infty} \frac{f(x)}{g(x)}=1$ you get $$\lim\limits_{x\to\infty} \frac{f(x)-g(x)}{g(x)}=0.$$

For example, if you have additional information that the limit $\lim\limits_{x\to\infty} g(x)=L$ exists and is finite, then you get $$\lim\limits_{x\to\infty} (f(x)-g(x)) = \lim\limits_{x\to\infty} \frac{f(x)-g(x)}{g(x)}\cdot g(x) = 0\cdot L = 0.$$

In fact, even a bit weaker assumption that $g(x)$ is bounded is enough. Take any $\varepsilon>0$. You have $|g(x)|\le M$ for some real number $M$ and each $x$ and $$\left|\frac{f(x)-g(x)}{g(x)}\right| \le \varepsilon$$ for every large enough $x$. Then you get $$|f(x)-g(x)| = \left|\frac{f(x)-g(x)}{g(x)}\right| \cdot |g(x)| \le \varepsilon M$$ for $x\ge x_0$ (where $x_0$ depends on $\varepsilon$), i.e., $$\limsup_{x\to\infty} |f(x)-g(x)| \le \varepsilon M.$$ Since this is true for every $\varepsilon > 0$ (and $M$ is fixed), you get $\limsup\limits_{x\to\infty} |f(x)-g(x)| = 0$ and $$\lim\limits_{x\to\infty} (f(x)-g(x)) = 0.$$