It may be useful to construct the relevant distributions the other way round, that is, to start from a single exponential random variable $X_\lambda$ with density $\lambda\mathrm e^{-\lambda x}$ on $x\geqslant0$, for a given positive $\lambda$, and to consider the integer part $Y_a$ of $X_\lambda/a$, for every positive $a$.
Then, $[Y_a=n]=[na\leqslant X_\lambda\lt (n+1)a]$ hence $\mathrm P(Y_a=n)=\mathrm e^{-\lambda na}-\mathrm e^{-\lambda (n+1)a}$ for every nonnegative integer $n$, that is, $\mathrm P(Y_a=n)=(p_a)^n(1-p_a),$ with $ p_a=\mathrm e^{-\lambda a}. $ This proves that each $Y_a$ is geometric with parameter $p_a$. Note that, for a given $\lambda$, every parameter $p$ in $(0,1)$ is realized as $p=p_a$ for some $a$, hence, starting from a unique random variable $X_\lambda$, this construction yields a whole family of geometric random variables $(Z_p)_{p\in(0,1)}$ such that each $Z_p$ is geometric with parameter $p$ and such that $Z_p\leqslant Z_q$ with full probability, for every $p\geqslant q$ in $(0,1)$: simply define each $Z_p$ by $Z_p=Y_{\alpha(p)}$, with $ \alpha(p)=-\log(p)/\lambda. $ Coming back to your question, $X_\lambda\leqslant aY_a\lt X_\lambda+a$ with full probability, hence $aY_a\to X_\lambda$ almost surely when $a\to0$, in particular $\alpha(p)Z_p\to X_\lambda$ when $p\to1$. Thus, $-\log(p)Z_p$ converges almost surely (hence in distribution) to the standard exponential random variable $X_1=\lambda X_\lambda$.
Since $-\log(p)\sim1-p$ when $p\to1$, this shows that $(1-p)Z_p$ converges almost surely (hence also in distribution) to $X_1$. Now, the convergence in distribution of some random variables does not depend on their realization hence, by the remarks above, for every family $(T_p)_{p\in(0,1)}$ of random variables such that each $T_p$ is geometrically distributed with parameter $p$, the random variables $(1-p)T_p$ converge in distribution to the standard exponential distribution when $p\to1$.
Note Such a simultaneous and almost sure construction of geometric random variables $Z_p$, for every $p$ in $(0,1)$, from a single (exponentially distributed) random variable $X_\lambda$ is called a coupling. For more about this powerful idea, see the WP page on coupling in probability.